Explanation

(a)
Given that, $x = a\cos^{4}\theta and\ y = a\sin^{4}\theta$
On differentiating w.r.t. $\theta$, we get
$\frac{dx}{d\theta} = 4{a\cos}^{3}{\theta( - \sin{\theta)}}$
and $\frac{dy}{d\theta} = 4{a\sin}^{3}{\theta\cos\theta}$
$\therefore\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = - \frac{4a\sin^{3}{\theta\cos\theta}}{4{a\cos}^{3}{\theta\sin\theta}}$
$= - \frac{\sin^{2}\theta}{\cos^{2}\theta} = - \tan^{2}\theta$
Now, $\left( \frac{dy}{dx} \right)_{\theta = \frac{3\pi}{4}} = - \tan^{2}\left( \frac{3\pi}{4} \right) = - 1$