$Letx\left(t\right)=2\sqrt{2}\cos t\sqrt{\sin 2t}andy\left(t\right)=2\sqrt{2}\sin t\sqrt{\sin 2t},t\in (0,\frac{\pi }{2})then\frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}att=\frac{\pi }{4}isequalto$

$\begin{array}{l}\therefore \frac{dx}{dt}=\frac{2\sqrt{2}\cos 3t}{\sqrt{\sin 2t}}.\frac{dy}{dt}=\frac{2\sqrt{2}\sin 3t}{\sqrt{\sin 2t}}\\ \therefore \frac{dy}{dx}=\tan 3t.(att=\frac{\pi }{4},\frac{dy}{dx}=-1)\\ and\frac{d^{2}y}{d{x}^2}=3{\sec }^{2}3t.\frac{dt}{dx}=\frac{3{\sec }^{2}3t.\sqrt{\sin 2t}}{2\sqrt{2}\cos 3t}\\ (Att=\frac{\pi }{4},\frac{d^{2}y}{d{x}^2}=-3)\\ \therefore \frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}=\frac{2}{-3}=\frac{-2}{3}\end{array}$