$$\begin{gathered} Let x \left(t\right)=2\sqrt{2}\cos t \sqrt{\sin 2t}and \\ y \left(t\right)=2\sqrt{2}\sin t \sqrt{\sin 2t,}t\in \left(0,\frac{\mathrm{\pi }}{2}\right).Then \\ \frac{1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}{{\displaystyle \frac{d^{2}y}{d{x}^2}}}at t=\frac{\mathrm{\pi }}{4}is equal to \end{gathered}$$

$$\begin{gathered} x =2 \sqrt{2}\cos t \sqrt{\sin 2t }\Rightarrow \frac{dx}{dt}=2 \sqrt{2}\cos t\left(\frac{2\cos 2t}{2\sqrt{\sin 2t }}\right)+\sqrt{\sin 2t }\left(-2 \sqrt{2}\sin t\right) \\ \Rightarrow \frac{dx}{dt}=\frac{2\sqrt{2}\cos 3t}{\sqrt{\sin 2t}} \\ y\left(t\right)=2\sqrt{2}\sin t\sqrt{\sin 2t}\Rightarrow \frac{dy}{dt}=2\sqrt{2}\sin t\left(\frac{2\cos 2t}{2\sqrt{\sin 2t }}\right)+\sqrt{\sin 2t}\left(2\sqrt{2}\cos t\right) \\ \Rightarrow \frac{dy }{dt}=\frac{2\sqrt{2}\sin 3t}{\sqrt{\sin 2t}} \\ \frac{dy}{dx}=\tan 3t \\ \frac{dy}{dx}=-1 at t\frac{\mathrm{\pi }}{4} \\ \frac{d^{2}y}{d{x}^2}=\frac{3}{2\sqrt{2}}se{c}^3 3t.\sqrt{\sin 2t}=-3at t=\frac{\mathrm{\pi }}{4} \\ \therefore \frac{1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}{{\displaystyle \frac{d^{2}y}{d{x}^2}}}=\frac{1+1}{-3}=-\frac{2}{3} \end{gathered}$$