If $x=3\cos \theta -2{\cos }^3\theta$ and $y=3\sin \theta -2{\sin }^3\theta$, then ${\left(\frac{dy}{dx}\right)}_{\theta =\frac{\pi }{3}}=$

$x=3\cos \theta -2{\cos }^3\theta$

$\frac{dx}{d\theta }=3(-\sin \theta )-6{\cos }^2\theta (-\sin \theta )$

$=-3\sin \theta +6{\cos }^2\theta .\sin \theta$

$y=3\sin \theta -2{\sin }^3\theta$

$\frac{dy}{d\theta }=3\cos \theta -6{\sin }^2\theta .\cos \theta$

$\frac{dy}{dx}=\frac{3\cos \theta -6{\sin }^2\theta .\cos \theta }{6{\cos }^2\theta .\sin \theta -3\sin \theta }$

${\left(\frac{dy}{dx}\right)}_{\theta =\frac{\pi }{3}}=\frac{3\cos \frac{\pi }{3}-6.{\sin }^2\frac{\pi }{3}.\cos \frac{\pi }{3}}{6{\cos }^2\frac{\pi }{3}.\sin \frac{\pi }{3}-3\sin \frac{\pi }{3}}$

$\Rightarrow \frac{3.\frac{1}{2}-6.\frac{3}{4}.\frac{1}{2}}{6.\frac{1}{4}.\frac{\sqrt{3}}{2}-3.\frac{\sqrt{3}}{2}}=\frac{\frac{3}{2}-\frac{9}{4}}{\frac{3\sqrt{3}}{4}-\frac{3\sqrt{3}}{2}}$

$=\frac{1}{\sqrt{3}}$