Explanation

(b)
Given, $y = a\sin^{3}{\theta\ \ \ \ and\ }x = a\cos^{3}\theta$
On differentiating w.r.t. $\theta,\$we get
$\frac{dy}{d\theta}$ = $3a\sin^{2}\theta\cos\theta$
and $\frac{dx}{d\theta}$ = $- 3a\cos^{2}{\theta\sin\theta}$
$\therefore\ \ \ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = - \frac{3a\sin^{2}{\theta\cos\theta}}{3a\cos^{2}\theta\sin\theta} = {- tan}\theta$
At $\theta = \frac{\pi}{3},\frac{dy}{dx} = - \tan{\frac{\pi}{3} = - \sqrt{3}}$