$x = \cfrac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},y = \cfrac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$

Here $x = \cfrac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }}$ …(i) and $y = \cfrac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$ …(ii)

Differentiating (i) \& (ii) w.r.t. t, we get

$\cfrac{{dx}}{{dt}} = \cfrac{{\sqrt {\cos 2t} \cfrac{d}{{dt}}({{\sin }^3}t) - {{\sin }^3}t\cfrac{d}{{dt}}(\sqrt {\cos 2t} )}}{{\cos 2t}}$

$= \cfrac{{\sqrt {\cos 2t} 3{{\sin }^2}t\cos t - {{\sin }^3}t \cdot \cfrac{1}{{2\sqrt {\cos 2t} }} \cdot ( - \sin 2t) \cdot 2}}{{\cos 2t}}$

$= \cfrac{{\sqrt {\cos 2t} 3{{\sin }^2}t\cos t + \cfrac{{{{\sin }^3}t\sin 2t}}{{\sqrt {\cos 2t} }}}}{{\cos 2t}}$

$= \cfrac{{3\cos 2t{{\sin }^2}t\cos t + {{\sin }^3}t\sin 2t}}{{{{(\cos 2t)}^{3/2}}}}$

$\cfrac{{dy}}{{dt}} = \cfrac{{\sqrt {\cos 2t} \cfrac{d}{{dt}}({{\cos }^3}t) - {{\cos }^3}t\cfrac{d}{{dt}}(\sqrt {\cos 2t} )}}{{\cos 2t}}$

$= \cfrac{{\sqrt {\cos 2t} \cdot 3{{\cos }^2}t( - \sin t) - {{\cos }^3}t \cdot \cfrac{1}{{2\sqrt {\cos 2t} }} \cdot ( - \sin 2t) \cdot 2}}{{\cos 2t}}$

$= \cfrac{{ - 3{{\cos }^2}t \cdot \sin t \cdot \sqrt {\cos 2t} + \cfrac{{{{\cos }^3}t\sin 2t}}{{\sqrt {\cos 2t} }}}}{{\cos 2t}}$

$= \cfrac{{{{\cos }^3}t\sin 2t - 3{{\cos }^2}t \cdot \sin t\cos 2t}}{{{{(\cos 2t)}^{3/2}}}}$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{{{\cos }^3}t\sin 2t - 3{{\cos }^2}t \cdot \sin t\cos 2t}}{{3\cos 2t{{\sin }^2}t\cos t + {{\sin }^3}t\sin 2t}} = - \cot 3t$