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

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

therefore,$\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}(3\cos \theta ) - \frac{d}{{d\theta }}\left( {2{{\cos }^3}\theta } \right)$
$= 3 \cdot ( - \sin \theta ) - 2 \cdot 3{\cos ^2}\theta \cdot \frac{d}{{d\theta }} \cdot \cos \theta$

$= - 3\sin \theta + 6{\cos ^2}\theta \sin \theta$
and $\frac{{dy}}{{d\theta }} = 3\cos A - 2 \cdot 3{\sin ^2}\theta \cdot \frac{d}{{d\theta }} \cdot \sin \theta$

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

Now, $\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{3\cos \theta - 6{{\sin }^2}\theta \cos \theta }}{{ - 3\sin \theta + 6{{\cos }^2}\theta \sin \theta }}$

$= \frac{{3\cos \theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{3\sin \theta \left( { - 1 + 2{{\cos }^2}\theta } \right)}} = \cot \theta \cdot \frac{{\cos 2\theta }}{{\cos 2\theta }} = \cot \theta$

$\Rightarrow \frac{dy}{dx} = \cot \theta$