If $ x=\frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}\;,$ Then $ \frac{dy}{dx}$ in terms of $t$ If $\displaystyle x=\frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $\displaystyle y = \frac{\cos^3 t}{\sqrt{\cos 2t}}\;,$ Then $\displaystyle \frac{dy}{dx}$ in terms of $t$
$\bf{My\; Try::}$ Using The Formula $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$
Now Given $$\displaystyle x= \frac{\sin^3 t}{\sqrt{\cos 2t}}.$$ So $$\displaystyle \frac{dx}{dt} = \frac{\sqrt{\cos 2t}\cdot 3\sin^2 t \cdot \cos t-\sin^3 t \cdot \frac{1}{2\sqrt{\cos 2t}}\cdot -\sin 2t \cdot 2t}{\cos 2t}$$
So we get $$\displaystyle \frac{dx}{dt} = \frac{\cos 2t \cdot 3\sin^2 t \cdot \cos t+\sin^3 t \cdot \sin 2t \cdot t}{\cos 2t \cdot \sqrt{\cos 2t}}$$
Similarly Given $$\displaystyle y=\frac{\cos^3 t}{\sqrt{\cos 2t}}.$$ So $$\displaystyle \frac{dy}{dt} = \frac{-\sqrt{\cos 2t}\cdot 3\cos^2 t \cdot \sin t-\cos^3 t\cdot \frac{1}{2\sqrt{\cos 2t}}\cdot -\sin 2t \cdot 2t}{\cos 2t}$$
So we get $$\displaystyle \frac{dy}{dt} = \frac{-\cos 2t \cdot 3\cos^2 t \cdot \sin t+\cos^3 t\cdot \sin 2t\cdot t}{\cos 2t \cdot \sqrt{\cos 2t}}$$
So $$\displaystyle \frac{dy}{dx} = \frac{-3\cos^2 t \cdot \sin t\cdot \cos 2t+\cos^3 t\cdot \sin 2t\cdot t}{\cos 2t \cdot 3\sin^2 t \cdot \cos t+\sin^3 t \cdot \sin 2t \cdot t}$$
But this method seems complicated.
Is there is any other method by which we can calculate $\displaystyle \frac{dy}{dx}$?

You could simplify the intermediate calculations using reduction formulas of the powers of sine and cosine functions $$x=\frac{\sin ^3(t)}{\sqrt{\cos (2 t)}}$$ $$\frac{dx}{dt}=\frac{\sin (2 t) \sin ^3(t)}{\cos ^{\frac{3}{2}}(2 t)}+\frac{3 \sin ^2(t) \cos
   (t)}{\sqrt{\cos (2 t)}}=\frac{\sin ^2(t) (2 \cos (t)+\cos (3 t))}{\cos ^{\frac{3}{2}}(2 t)}$$ $$y=\frac{\cos ^3(t)}{\sqrt{\cos (2 t)}}$$ $$\frac{dy}{dt}=\frac{\sin (2 t) \cos ^3(t)}{\cos ^{\frac{3}{2}}(2 t)}-\frac{3 \sin (t) \cos
   ^2(t)}{\sqrt{\cos (2 t)}}=\frac{\sin (t)-\sin (5 t)}{4 \cos ^{\frac{3}{2}}(2 t)}$$ $$\frac{dy}{dx}=-\cot (3 t)$$
This was just done using your steps (I suppose some mistakes since $t$ terms cannot occur during the derivations). 
