Hints : $y=a \sin ^3 \mathrm{t} ; \mathrm{x}=\mathrm{a} \cos ^3 \mathrm{t}$
$$
\begin{aligned}
& \frac{\mathrm{dy}}{\mathrm{dt}}=3 \sin ^2 \mathrm{t} \cos \mathrm{t} ; \frac{\mathrm{dx}}{\mathrm{dt}}=-3 \mathrm{a} \cos ^2 \mathrm{t} \sin \mathrm{t} \\
& \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \operatorname{sint}}=-\frac{\operatorname{sint}}{\cos t}=-\operatorname{tant} \\
&
\end{aligned}
$$


$$
\begin{aligned}
& \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx} \mathrm{x}^2}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(-\tan t)=\frac{\mathrm{d}}{\mathrm{dt}}(-\operatorname{tant}) \cdot \frac{\mathrm{dt}}{\mathrm{dx}} \\
& =\left(-\sec ^2 \mathrm{t}\right) \frac{1}{-3 \cos ^2 \mathrm{t} \sin t}=\frac{1}{+3 \cos ^4 \mathrm{t} \sin t} \\
& \left.\frac{d^2 y}{d x^2}\right|_{t=\pi / 4}=\frac{1}{3 a\left(\frac{1}{\sqrt{2}}\right)^4 \cdot\left(\frac{1}{\sqrt{2}}\right)}=\frac{(\sqrt{2})^5}{3 a}=\frac{4 \sqrt{2}}{3 a}
\end{aligned}
$$