JEE Math Practice Question

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

1. A.$$\frac{-2 \sqrt{2}}{3}$$
2. B.$$\frac{2}{3}$$
3. C.$$\frac{1}{3}$$
4. D.$$ \frac{-2}{3}$$