Explanation

(d)
Given, $x = \cos{\theta,\ \ \ \ y = \sin{5\theta}}$
$\Rightarrow \ \frac{dx}{d\theta} = \ - \sin{\theta,\ \ \ \ \frac{dy}{d\theta} = 5\cos{5\theta}\ \ \ \ \ \ \ }$
$\therefore\ \ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = - \frac{5\cos{5\theta}}{\sin\theta}$
$\Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{d}{d\theta}\left( \frac{dy}{dx} \right).\frac{d\theta}{dx}$
$= \frac{d}{d\theta}\left( \frac{- 5\cos{5\theta}}{\sin\theta} \right)\frac{1}{- \sin\theta}\$
$= \left( \frac{\sin{\theta\sin{5\theta.25 + 5\cos{5\theta\cos\theta}}}}{\sin^{2}\theta} \right).\frac{1}{- \sin\theta}\$
$= - \frac{25\sin{5\theta}}{\sin^{2}\theta} - \frac{5\cos{5\theta\cos\theta}}{\sin^{3}\theta}$
$\therefore\left( 1 - x^{2} \right)\frac{d^{2}y}{dx^{2}} - x\frac{dy}{dx}$
$= (1 - \cos^{2}{\theta)}\left( \frac{- 25\sin{5\theta}}{\sin^{2}\theta} - \frac{5\cos{5\theta\cos\theta}}{\sin^{3}\theta} \right)$
$- \cos{\theta\left( \frac{- 5\cos{5\theta}}{\sin\theta} \right)}$
$= {\ sin}^{2}\theta\left( \frac{- 25\sin{5\theta}}{\sin^{2}\theta} - \frac{5\cos{\theta\cos{5\theta}}}{\sin^{3}\theta} \right) + \frac{5\cos{\theta\cos{5\theta}}}{\sin\theta}$
$= - 25\sin{5\theta - \frac{5\cos{\theta\cos{5\theta}}}{\sin\theta} + \frac{5\cos{\theta\cos{5\theta}}}{\sin\theta}}$
= $- 25y$