If \(x=a \cos ^{3} \theta\) and \(y=a \sin ^{3} \theta\), then \(\frac{d y}{d x}\) is

If \(x=a \cos ^{3} \theta\) and \(y=a \sin ^{3} \theta\)
\(\Rightarrow \quad \frac{d x}{d \theta}=3 a \cos ^{2} \theta(-\sin \theta)\)
\(\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{d} \theta}=3 \mathrm{a} \sin ^{2} \theta(\cos \theta)\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}\)
\(=\left(3 a \sin ^{2} \theta \cdot \cos \theta\right) \cdot \frac{1}{-\left(3 a \sin \theta \cdot \cos ^{2} \theta\right)}\)
\(\frac{d y}{d x}=-\frac{\sin \theta}{\cos \theta}, \frac{d y}{d x}=-\tan \theta\)
\(\Rightarrow \quad \frac{d y}{d x}=-\frac{(y / a)^{-1 / 3}}{(x / a)^{1 / 3}}\)
\[
\begin{aligned}
&=-\left[\frac{y}{a} \times \frac{a}{x}\right]^{1 / 3} \\
\frac{d y}{d x} &=-\left(\frac{y}{x}\right)^{-1 / 3}=-\sqrt[3]{\frac{y}{x}}
\end{aligned}
\]