If $=3\mathrm{cost}$ and $\mathrm{y}=4\mathrm{sint},$ then $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ at the point $\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)=\left(\frac{3}{2}\sqrt{2},2\sqrt{2}\right),$ is

$$\begin{gathered} \mathrm{x}=3\cos \mathrm{t}, \mathrm{y}=4\sin \mathrm{t} \\ \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{16}=1 \Rightarrow 9{\mathrm{y}}^2+16{\mathrm{x}}^2=144 \\ \frac{2\mathrm{x}}{9}+\frac{2\mathrm{y}}{16}\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-16}{9}\frac{\mathrm{x}}{\mathrm{y}} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{-16}{9} \left(\frac{\mathrm{y}-\mathrm{x}{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}}{{\mathrm{y}}^2}\right) \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{-16}{9}\left(\frac{\mathrm{y}+{\displaystyle \frac{16{\mathrm{x}}^2}{9\mathrm{y}}}}{{\mathrm{y}}^2}\right) \\ =\frac{-16}{9}\left(\frac{9{\mathrm{y}}^2+16{\mathrm{x}}^2}{9{\mathrm{y}}^2}\right) \\ =\frac{-16}{81{\mathrm{y}}^3}\left(144\right) \\ {\left(\frac{{\mathrm{d}}^{2\mathrm{y}}}{{\mathrm{dx}}^2}\right)}_{\left(\frac{3}{2}\sqrt{2}, 2\sqrt{2}\right)}=\frac{-16\times 144}{81{\left(2\sqrt{2}\right)}^3} \\ = \frac{-16\times 16\times 9}{81 16\sqrt{2}} \\ =\frac{-8\sqrt{2}}{9} \end{gathered}$$