We are given that $x=a \cos ^3 \theta, y=a \sin ^3 \theta$
Now $\frac{d x}{d \theta}=3 a \cos ^2 \theta(-\sin \theta), \frac{d y}{d \theta}=3 a \sin ^2 \theta(\cos \theta)$
Now, $\frac{d y}{d x}=\frac{3 a \sin ^2 \theta \cos \theta}{-3 a \cos ^2 \theta \sin \theta}=-\tan \theta$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}=-\tan \theta \Rightarrow \frac{d y}{d x^2}=-\sec ^2 \theta \times\left(\frac{d \theta}{d x}\right) \\ & \Rightarrow \frac{d^2 y}{d x^2}=-\sin ^2 \theta \times \frac{1}{-30 \cos ^2 \theta \times \sin \theta} \\ & \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{1}{39}\right) \frac{1}{\sin \theta \times \cos ^4 \theta}\end{aligned}$
At $\theta=\frac{\pi}{4}$
$\begin{aligned} & \Rightarrow \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left(\frac{1}{39}\right) \frac{1}{\sin \left(\frac{\pi}{4}\right)\left(\cos \frac{\pi}{4}\right)^4} \\ & \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left(\frac{1}{39}\right) \times \frac{1}{\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2 \times 2}\right)}=\frac{4 \sqrt{2}}{39}\end{aligned}$