$x=\sin \theta \sqrt{\cos \theta },y=\cos \theta \sqrt{\cos 2\theta }\text{ then the value of }\frac{dy}{dx}\text{ at }\theta =\frac{\pi }{4}\text{ is }$

$\frac{dx}{d\theta }=\sin \theta \frac{1}{2\sqrt{\cos \theta }}\left(-\sin \theta \right)+\sqrt{\cos \theta }\cos \theta -----\left(1\right)$

$\begin{array}{l}\frac{dy}{d\theta }=\cos \theta \frac{2}{2\sqrt{\cos 2\theta }}(-\sin 2\theta )+\sqrt{\cos 2\theta }(-\sin \theta )\\ =\frac{-\cos \theta 2\sin \theta \cos \theta }{\sqrt{\cos 2\theta }}-\sin \theta \sqrt{\cos 2\theta }\\ \frac{dy}{d\theta }=\frac{-2{\cos }^2\theta \sin \theta -\sin \theta \cos 2\theta }{\sqrt{\cos 2\theta }}\\ \frac{dy}{dx}=\frac{dy\mid d\theta }{dx\mid d\theta }=\frac{\frac{-2{\cos }^2\theta \sin \theta -\sin \theta \cos 2\theta }{\sqrt{\cos 2\theta }}}{\frac{-{\sin }^2\theta }{2\sqrt{\cos \theta }}+\sqrt{\cos \theta }\cos \theta }\end{array}$

$\theta =\frac{\pi }{4},\frac{dy}{dx}=-2\frac{1}{2}\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\left(0\right)=0$