Let $f(x)=\mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{{n^n}(x + n)(x + \dfrac{n}{2}).....(x + \dfrac{n}{n})}}{{n!({x^2} + {n^2})({x^2} + \dfrac{{{n^2}}}{4}).....({x^2} + \dfrac{{{n^2}}}{{{n^2}}})}}} \right)^{\dfrac{x}{n}}}$, for all $x > 0$. Then:
A) $f(\dfrac{1}{2}) \geqslant f(1) \\$
B) $f(\dfrac{1}{3}) \leqslant f(\dfrac{2}{3}) \\$
C) ${f'}(2) \leqslant 0 \\$
D) $\dfrac{{{f'}(3)}}{{f(3)}} \geqslant \dfrac{{{f'}(2)}}{{f(2)}} \\$

Hint: In this differential equation problem, first of all, we shall have to simplify the expression given by taking out common factors cancelling them. After that we will put the values given in the options to check which option (s) is (are) correct.

Complete step by step answer:
Firstly, rewrite the expression given in the question in the following way,
$f(x) = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{{n^n}(x + n)(x + \dfrac{n}{2}).....(x + \dfrac{n}{n})}}{{n!({x^2} + {n^2})({x^2} + \dfrac{{{n^2}}}{4}).....({x^2} + \dfrac{{{n^2}}}{{{n^2}}})}}} \right)^{\dfrac{x}{n}}}$
Now, from the numerator, take out $n$ from all bracketed terms and ${n^2}$ from each bracketed terms in denominator, we will get the following expression,
$f(x) = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{{n^n}.{n^n}(\dfrac{x}{n} + 1)(\dfrac{x}{n} + \dfrac{1}{2}).....(\dfrac{x}{n} + \dfrac{1}{n})}}{{n!.{{({n^2})}^n}(\dfrac{{{x^2}}}{{{n^2}}} + 1)(\dfrac{{{x^2}}}{{{n^2}}} + \dfrac{1}{4}).....(\dfrac{{{x^2}}}{{{n^2}}} + \dfrac{1}{{{n^2}}})}}} \right)^{\dfrac{x}{n}}}$
Now, simplifying for $n$, we get,
$f(x) = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{{n^{2n}}(\dfrac{x}{n} + 1)(\dfrac{x}{n} + \dfrac{1}{2}).....(\dfrac{x}{n} + \dfrac{1}{n})}}{{n!.{n^2}^n(\dfrac{{{x^2}}}{{{n^2}}} + 1)(\dfrac{{{x^2}}}{{{n^2}}} + \dfrac{1}{4}).....(\dfrac{{{x^2}}}{{{n^2}}} + \dfrac{1}{{{n^2}}})}}} \right)^{\dfrac{x}{n}}}$
In the above,${n^{2n}}$ term will be cancelled out. Now, in the denominator, we know that$n! = 1.2.3.4.5..............n$, so we multiply 1 with first term, 2 with second term, and so on. We will get following expression,
 \[
  f(x) = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{{n^{2n}}(\dfrac{x}{n} + 1)(\dfrac{x}{n} + \dfrac{1}{2}).....(\dfrac{x}{n} + \dfrac{1}{n})}}{{n!.{n^2}^n(\dfrac{{{x^2}}}{{{n^2}}} + 1)(\dfrac{{{x^2}}}{{{n^2}}} + \dfrac{1}{4}).....(\dfrac{{{x^2}}}{{{n^2}}} + \dfrac{1}{{{n^2}}})}}} \right)^{\dfrac{x}{n}}} \\
  f(x) = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{(\dfrac{x}{n} + 1)(\dfrac{x}{n} + \dfrac{1}{2}).....(\dfrac{x}{n} + \dfrac{1}{n})}}{{(1.\dfrac{{{x^2}}}{{{n^2}}} + 1)(2.\dfrac{{{x^2}}}{{{n^2}}} + 2.\dfrac{1}{4}).....(n.\dfrac{{{x^2}}}{{{n^2}}} + n.\dfrac{1}{{{n^2}}})}}} \right)^{\dfrac{x}{n}}} \\
  f(x) = \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{(\dfrac{x}{n} + 1)(\dfrac{x}{n} + \dfrac{1}{2}).....(\dfrac{x}{n} + \dfrac{1}{n})}}{{(\dfrac{{{x^2}}}{{{n^2}}} + 1)(\dfrac{{2{x^2}}}{{{n^2}}} + \dfrac{1}{2}).....(\dfrac{{n{x^2}}}{{{n^2}}} + \dfrac{1}{n})}}} \right)^{\dfrac{x}{n}}} \\
 \]
Now, taking log on both sides, we get the following expression,
\[
  \log (f(x)) = \log \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{(\dfrac{x}{n} + 1)(\dfrac{x}{n} + \dfrac{1}{2}).....(\dfrac{x}{n} + \dfrac{1}{n})}}{{(\dfrac{{{x^2}}}{{{n^2}}} + 1)(\dfrac{{2{x^2}}}{{{n^2}}} + \dfrac{1}{2}).....(\dfrac{{n{x^2}}}{{{n^2}}} + \dfrac{1}{n})}}} \right)^{\dfrac{x}{n}}} \\
  \log (f(x)) = \mathop {\lim }\limits_{n \to \infty } \dfrac{x}{n}[\left( {\log (\dfrac{x}{n} + 1) + \log (\dfrac{x}{n} + \dfrac{1}{2}) + ..... + \log (\dfrac{x}{n} + \dfrac{1}{n})} \right) \\
  {\text{ }} - \left( {\log (\dfrac{{{x^2}}}{{{n^2}}} + 1) + \log (\dfrac{{2{x^2}}}{{{n^2}}} + \dfrac{1}{2}) + ..... + \log (\dfrac{{n{x^2}}}{{{n^2}}} + \dfrac{1}{n})} \right)] \\
 \]
Now, we get two series in terms of \[\log (\dfrac{x}{n} + 1)\] and \[\log (\dfrac{{{x^2}}}{{{n^2}}} + 1)\]. We can write these series in summation form as below,
\[\log (f(x)) = \mathop {\lim }\limits_{n \to \infty } \dfrac{x}{n}[\sum\limits_{r = 1}^n {\log (\dfrac{x}{n} + \dfrac{1}{r})} - \sum\limits_{r = 1}^n {\log (\dfrac{{r{x^2}}}{{{n^2}}} + \dfrac{1}{r})} ]\]
This can be re-written as by taking summation as common, because the summation range is same for both expressions,
\[\log (f(x)) = \mathop {\lim }\limits_{n \to \infty } \dfrac{x}{n}[\sum\limits_{r = 1}^n {\left( {\log (\dfrac{x}{n} + \dfrac{1}{r}) - \log (\dfrac{{r{x^2}}}{{{n^2}}} + \dfrac{1}{r})} \right)} ]\],
 Now taking out $\dfrac{1}{r}$ common from both terms, we get,
\[\log (f(x)) = \mathop {\lim }\limits_{n \to \infty } \dfrac{x}{n}[\sum\limits_{r = 1}^n {\left( {\log \dfrac{1}{r}(\dfrac{{rx}}{n} + 1) - \log \dfrac{1}{r}(\dfrac{{{r^2}{x^2}}}{{{n^2}}} + 1)} \right)} ]\]
This can be written as
\[
  \log (f(x)) = \mathop {\lim }\limits_{n \to \infty } \dfrac{x}{n}[\sum\limits_{r = 1}^n {\left( {\log \dfrac{1}{r} + \log (\dfrac{{rx}}{n} + 1) - \log \dfrac{1}{r} - \log (\dfrac{{{r^2}{x^2}}}{{{n^2}}} + 1)} \right)} ] \\
  \log (f(x)) = \mathop {\lim }\limits_{n \to \infty } \dfrac{x}{n}[\sum\limits_{r = 1}^n {\left( {\log (\dfrac{{rx}}{n} + 1) - \log (\dfrac{{{r^2}{x^2}}}{{{n^2}}} + 1)} \right)} \\
 \]
Assuming that $\dfrac{r}{n} = y,$then the above equation can be written as
$\log (f(x)) = \mathop {\lim }\limits_{n \to \infty } x[\sum\limits_{r = 1}^n {\left( {\log (xy + 1) - \log ({y^2}{x^2} + 1)} \right)} $
Now, changing it into the integration form as
$\log (f(x)) = x\int\limits_0^1 {\log (1 + xy)dy - } x\int\limits_0^1 {\log (1 + {x^2}{y^2})dy} $
Now, let
$
  xy = t \\
  {{Then, }}xdy = dt \\
$
So that the above equation will become, and range will become 0 to$x$,
 $\log (f(x)) = \int\limits_0^x {\log (1 + t)dt - } \int\limits_0^x {\log (1 + {t^2})dt} $
After simplification, we get
$\log (f(x)) = \int\limits_0^x {\log \dfrac{{(1 + t)}}{{(1 + {t^2})}}dt} $
Now differentiate, we get
$\dfrac{{{f'}(x)}}{{f(x)}} = \log \dfrac{{1 + x}}{{1 + {x^2}}}$
Now, from the options given, we will check these options one by one.
Now, put$x = 2$, we get
$\dfrac{{{f'}(2)}}{{f(2)}} = \log (\dfrac{{1 + 2}}{{1 + {2^2}}}) = \log (\dfrac{3}{5}) < 0$
It means, ${f'}(2) < 0$

Therefore, option (C) is correct.

Now, putting$x = 3$, we get
$\dfrac{{{f'}(3)}}{{f(3)}} = \log (\dfrac{{1 + 3}}{{1 + {3^2}}}) = \log \dfrac{4}{{10}} = \log (\dfrac{2}{5}) < 0$
But, this value is less than$\dfrac{{{f'}(2)}}{{f(2)}}$
Therefore, option (D) is incorrect.

Now, from the above trend, it can be said that for all$x > 0$
$\log \dfrac{{1 + x}}{{1 + {x^2}}}$> 0. Thus the function is an increasing function.
Now, we know that $\dfrac{1}{2} < 1$therefore, for this function, we can say that
$f(\dfrac{1}{2}) < f(1)$

Therefore, option (A) is incorrect.

Similarly, we know that $\dfrac{1}{3} < \dfrac{2}{3}$therefore, for this function, we can say that
$f(\dfrac{1}{3}) < f(\dfrac{2}{3})$.

Therefore, option (B) is correct.

This way, options (B) and (C) are the correct options.

Note: It is very common practice to solve these types of series and differential equation problems by cancelling the common factors and doing the substitution to reduce the complexity of the equation. These equations, though, look very difficult to get the answers from them, but if you see the above step by step procedure, these can be solved. You shall have to identify which type of substitution is required to simplify the equation.