The question given to me was actually of parametric differentiation, and the equations were:
$$x = \dfrac{\sin^3 t}{\sqrt{\cos2t}}\ , \ \ \ \ y = \dfrac{\cos^3 t}{\sqrt{\cos2t}}$$
and we had to find the $\dfrac{dy}{dx}$
One way of calculating could be calculating $\dfrac{dy}{dt}$ and dividing it by $\dfrac{dx}{dt}$.
But this method was very long and tiring, and takes time to solve. I was thinking of eliminating the parameter, but couldn't do so. Expressing $\cos t$ as $\sin(\frac{\pi}{2}-t)$ also doesn't help!
Is there a way by which we can eliminate the parameter $t$ ?
\begin{align*} \dot{x} &= \frac{3\sin^{2} t \cos t}{\sqrt{\cos 2t}}+ \frac{\sin^{3} t \sin 2t}{(\cos 2t)^{3/2}} \\ &=\frac{\sin^{2} t}{(\cos 2t)^{3/2}} ( 3\cos t \cos 2t+\sin t \sin 2t) \\ &=\frac{\sin^{2} t}{(\cos 2t)^{3/2}} (3\cos t \cos 2t+2\sin^{2} t \cos t) \\ &=\frac{\sin^{2} t \cos t}{(\cos 2t)^{3/2}} (3\cos 2t+2\sin^{2} t) \\ \dot{y} &=- \frac{3\cos^{2} t \sin t}{\sqrt{\cos 2t}}+ \frac{\cos^{3} t \sin 2t}{(\cos 2t)^{3/2}} \\ &=\frac{\cos^{2} t}{(\cos 2t)^{3/2}} (-3\sin t \cos 2t+\cos t \sin 2t) \\ &=\frac{\cos^{2} t}{(\cos 2t)^{3/2}} (-3\sin t \cos 2t+2\cos^{2} t \sin t) \\ &=\frac{\cos^{2} t \sin t}{(\cos 2t)^{3/2}} (-3\cos 2t+2\cos^{2} t) \\ \frac{dy}{dx} &= \frac{\cos t(2\cos^{2} t-3\cos 2t)}{\sin t(2\sin^{2} t+3\cos 2t)} \\ &=\frac{\cos t[2\cos^{2} t-3(2\cos^{2} t-1)]} {\sin t[2\sin^{2} t+3(1-2\sin^{2} t)]} \\ &=-\frac{4\cos^{3} t-3\cos t}{3\sin t-4\sin^{3} t} \\ &=-\frac{\cos 3t}{\sin 3t} \\ &=-\cot 3t \end{align*}
Further points to be noticed
\begin{align*} y^{2}-x^{2} &= \frac{7+\cos 4t}{8} \\ x^{2/3}+y^{2/3} &= \frac{1}{\sqrt[3]{\cos 2t}} \\ y^{2}-x^{2} &= \frac{1}{4} \left[ 3+\frac{1}{(x^{2/3}+y^{2/3})^{6}} \right] \\ \frac{dy}{dx} &=-\frac{3\cot t-\cot^{3} t}{1-3\cot^{2} t} \\ &=\frac{xy^{3}-3\sqrt[3]{x^{5}y}}{x^{2}-3y^{2}} \end{align*}