SOLUTION —
Given that, $y=\cos ^{-1} \sqrt{1-t^2}=\sin ^{-1} t$
and $x=\sin ^{-1}\left(3 t-4 t^3\right)=3 \sin ^{-1} t$
$\therefore \quad \frac{d y}{d t}=\frac{1}{\sqrt{1-t^2}}$
and $\frac{d x}{d t}=\frac{3}{\sqrt{1-t^2}}$
$\therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\left(\frac{1}{\sqrt{1-t^2}}\right)}{3\left(\frac{1}{\sqrt{1-t^2}}\right)}=\frac{1}{3}$
So, The correct option will be (D).