The equation is given as,

$y={\sin }^2\left({\cot }^{-1}\sqrt{\frac{1+x}{1-x}}\right)$

Substitute $x=\cos 2\theta \dots \left(\mathrm{I}\right)$

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

$y={\sin }^2\left({\cot }^{-1}\sqrt{\frac{2{\cos }^2\theta }{2{\sin }^2\theta }}\right)$

$y={\sin }^2\left({\cot }^{-1}\left(\cot \theta \right)\right)$

$y={\sin }^2\theta \dots \left(\mathrm{II}\right)$

Differentiate the above equation with respect to $\theta$,

$\frac{dy}{d\theta }=2\sin \theta \cos \theta$

$\frac{dy}{d\theta }=\sin 2\theta$

Differentiate the above equation (I) with respect to $\theta$,

$\frac{dx}{d\theta }=-2\sin 2\theta$

From equation $\left(\mathrm{III}\right) \text{and} \left(\mathrm{IV}\right)$

$\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{\sin 2\theta }{-2\sin 2\theta }$

$\frac{dy}{dx}=\frac{-1}{2}$