How do you differentiate ${{x}^{2}}y+x{{y}^{2}}=6$?

Hint: Now to differentiate the given equation we will first expand the differential by using $\dfrac{d\left( f+g \right)}{dx}=\dfrac{df}{dx}+\dfrac{dg}{dx}$ . Now we know that differentiation of constant is 0. Further we will expand the differential by using the product rule of differentiation which states $\dfrac{d\left( f\left( x \right).g\left( x \right) \right)}{dx}=f\left( x \right)\dfrac{d\left( g\left( x \right) \right)}{dx}+g\left( x \right)\dfrac{d\left( f\left( x \right) \right)}{dx}$ . Now we will use chain rule to differentiate terms in y. We will further simplify the expression and hence we have the differentiation of the given equation.

Complete step-by-step solution:
Now let us first understand the concept of differentiation.
Differentiation is nothing but rate of change of a function with respect to independent variable.
Differentiation of y with respect to x is denoted by $\dfrac{dy}{dx}$
Now if we have two functions in addition then we have $\dfrac{d\left( f+g \right)}{dx}=\dfrac{df}{dx}+\dfrac{dg}{dx}$
Now let us understand the chain rule of differentiation.
According to product rule of differentiation we have $\dfrac{d\left( f\left( x \right).g\left( x \right) \right)}{dx}=f\left( x \right)\dfrac{d\left( g\left( x \right) \right)}{dx}+g\left( x \right)\dfrac{d\left( f\left( x \right) \right)}{dx}$
Hence with the help of chain rule we can differentiate composite function.
Now consider the given equation ${{x}^{2}}y+x{{y}^{2}}=6$.
Now differentiating the equation on both sides with respect to x we get,
$\Rightarrow \dfrac{d\left( {{x}^{2}}y+x{{y}^{2}} \right)}{dx}=\dfrac{d6}{dx}$
Now we know that the constant of the differentiation of constant is 0 hence we have $\dfrac{d6}{dx}=0$
Now using the addition rule of differentiation we get,
$\Rightarrow \dfrac{d\left( {{x}^{2}}y \right)}{dx}+\dfrac{d\left( x{{y}^{2}} \right)}{dx}=0$
Now using the product rule we have,
$\Rightarrow \dfrac{d{{x}^{2}}}{dx}y+\dfrac{dy}{dx}{{x}^{2}}+\dfrac{dx}{dx}{{y}^{2}}+\dfrac{d{{y}^{2}}}{dx}x=0$
Now we have the differentiation ${{x}^{n}}=n{{x}^{n-1}}$ and $\dfrac{d{{y}^{2}}}{dx}=\dfrac{d{{y}^{2}}}{dy}\dfrac{dy}{dx}$
Hence using this we get,
$\Rightarrow 2xy+{{x}^{2}}\dfrac{dy}{dx}+{{y}^{2}}+2xy\dfrac{dy}{dx}=0$
Hence differentiating the given equation we get, $2xy+{{x}^{2}}\dfrac{dy}{dx}+{{y}^{2}}+2xy\dfrac{dy}{dx}=0$

Note: Now consider note that if we have differentiation of a fraction then we have $\dfrac{d\left( \dfrac{f}{g} \right)}{dx}=\dfrac{f'g-g'f}{{{g}^{2}}}$ We can derive this function by considering the fractions as $f\left( x \right){{\left( g\left( x \right) \right)}^{-1}}$ and use the product rule to find the differentiation. Also note that to differentiate ${{\left( g\left( x \right) \right)}^{-1}}$ we will use the chain rule of differentiation.