If $ x = \cos ec\,\theta - \sin \,\theta $ and $ y = \cos e{c^n}\theta - {\sin ^n}\theta $ ,
Then show that $ \left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) = 0 $ .

Hint: $ {\left( {\dfrac{{dy}}{{dx}}} \right)^2} $ is not equal to $ \dfrac{{{d^2}y}}{{d{x^2}}} $ as $ {\left( {\dfrac{{dy}}{{dx}}} \right)^2} $ is the square of $ \dfrac{{dy}}{{dx}} $ and $ \dfrac{{{d^2}y}}{{d{x^2}}} $ is the double differentiation of $ y $ with respect to $ x $ . $ {\sin ^n}\theta $ on differentiation do not $ {\cos ^n}\theta $ . as it's totally different thing. It's $ {\left( {\sin \,\theta } \right)^n} $ which will be differentiated by the rule $ \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} $ method. The parametric form is a method in which you differentiate two values with same value and later divide them, for example
if $ x = \sin \theta $ and $ y = \cos \,\theta $ , Find $ \dfrac{{dy}}{{dx}} $
Thus, $ \dfrac{{dx}}{{d\theta }} = \cos \theta $ and $ \dfrac{{dy}}{{d\theta }} = - \sin \theta $
Thus, $ \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{ - \sin \theta }}{{\cos \theta }} = \tan \theta $

Complete step-by-step answer:
Given: $ \left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) = 0 $
First differentiate $ x $ and $ y $ with respect to $ \theta $ to get the values,
 $
  x = \cos ec\,\theta - \sin \theta \\
\Rightarrow \dfrac{{dx}}{{d\theta }} = - \cot \theta .\cos ec\,\theta - \cos \theta \\
  \,\,\,\,\,\,\, = \dfrac{{ - \cos \theta }}{{\sin \theta }} \times \cos ec\,\theta - \cos \theta \\
  \,\,\,\,\,\,\, = - \cos \theta \left( {\dfrac{{\cos ec\,\theta }}{{\sin \theta }} + 1} \right) \\
  \dfrac{{dx}}{{d\theta }} = - \dfrac{{\cos \theta }}{{\sin \theta }}\left( {\cos ec\,\theta + \sin \theta } \right)\,\,\,\,\,\, - - - - - - \left( i \right) \\
  $
Now, for $ y $ ,
\[
  y = \cos ec{\,^n}\theta - {\sin ^n}\theta \\
\Rightarrow \dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right) \\
  \,\,\,\,\,\,\, = n\cos ec{\,^{n - 1}}\theta \left( { - \cot \theta .\cos ec\,\theta } \right) - n{\sin ^{n - 1}}\theta .\cos \theta \\
  \,\,\,\,\,\,\, = - n\cot \theta \left( {\cos e{c^n}\,\theta + {{\sin }^{n - 1}}\theta .\sin \theta } \right) \\
  \dfrac{{dy}}{{d\theta }} = - n\cot \theta \left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)\,\,\,\,\,\, - - - - - - \left( {ii} \right) \\
 \]
Now to find $ \dfrac{{dy}}{{dx}} $ , first divide \[\dfrac{{dy}}{{d\theta }}\] and $ \dfrac{{dx}}{{d\theta }} $ .
 $
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{ - n\cot \theta \left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)}}{{ - \cot \theta \left( {\cos ec\,\theta + \sin \theta } \right)}} \\
  \dfrac{{dy}}{{dx}} = \dfrac{{n\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)}}{{\left( {\cos ec\,\theta + \sin \theta } \right)}} \\
  $
Now, putting the value of $ \dfrac{{dy}}{{dx}} $ in the given equation.
\Rightarrow $ \left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) = 0 $
Now, Left hand side is
 $
\Rightarrow \left( {{x^2} + y} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} - {n^2}\left( {{y^2} + 4} \right) \\
   = \left( {{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2} + 4} \right)\left( {{n^2}\dfrac{{{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2}}}{{{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2}}}} \right) - {n^2}\left( {{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2} + 4} \right) \\
   = \left( {\cos ec{\,^2}\theta + {{\sin }^2}\theta - 2 + 4} \right)\left( {{n^2}\dfrac{{{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2}}}{{{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2}}}} \right) - {n^2}\left( {\cos ec{\,^{2n}}\theta + {{\sin }^{2n}}\theta - 2 + 4} \right) \\
   = \left( {\cos ec{\,^2}\theta + {{\sin }^2}\theta + 2\cos ec\,\theta .\sin \theta } \right){n^2}{\left( {\dfrac{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}}{{\left( {\cos ec\,\theta - \sin \theta } \right)}}} \right)^2} - {n^2}\left( {\cos ec{\,^{2n}}\theta + {{\sin }^{2n}}\theta + 2\cos ec{\,^n}\theta .{{\sin }^n}\theta } \right) \\
   = {\left( {\cos ec\,\theta + \sin \theta } \right)^2}{n^2}\left( {\dfrac{{{{\left( {\cos ec{\,^n}\theta - {{\sin }^n}\theta } \right)}^2}}}{{{{\left( {\cos ec\,\theta - \sin \theta } \right)}^2}}}} \right) - {n^2}{\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)^2} \\
   = {n^2}{\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)^2} - {n^2}{\left( {\cos e{c^n}\,\theta + {{\sin }^n}\theta } \right)^2} \\
   = 0 \\
  $
Hence the left hand side proved equal to right hand side

Note: In this type of questions students often make mistakes while differentiating. Do not directly differentiate $ y $ with respect to $ x $ to get $ \dfrac{{dy}}{{dx}} $ as it will be completely incorrect, Hence try to use another variable ” $ \theta $ ” and then divide them to get $ \dfrac{{dy}}{{dx}} $ , Now perform the operation on the given equation very carefully. A simple error in sign can bring you out the wrong result.