If $y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ then $\frac{d y}{d x}=$

$-\frac{2}{1+x^2}$

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

Taking sin on both sides

$\Rightarrow \sin y=\frac{1-x^2}{1+x^2}$

so third side by pythagoras theorem

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

$=2 x$

differentiating sin y w.r.t x

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

$\cos y \frac{dy}{dx} = \frac{(1+x^2)\frac{d}{dx}(1-x^2)-(1-x^2)\frac{d}{dx}(1+x^2)}{(1+x^2)^2}$

$\Rightarrow \cos y \frac{d y}{d x} =\frac{\left(1+x^2\right)(-2 x)-\left(1-x^2\right)(2 x)}{\left(1+x^2\right)^2}$

$\cos y \frac{d y}{d x} =\frac{2 x\left(-1-x^2-1+x^2\right)}{\left(1+x^2\right)^2}$

$\cos y \frac{d y}{d x}=\frac{2 x(-2)}{\left(1+x^2\right)^2} \Rightarrow \frac{d y}{d x}= \frac{1}{\cos y} \frac{(-4 x)}{\left(1+x^2\right)^2}$

from pythagoras triangle 

$\Rightarrow \cos y=\frac{2 x}{1+x^2}$

so  $\frac{d y}{d x}=\frac{\left(1+x^2\right)}{2 x} \frac{(-4 x)}{\left(1+x^2\right)^2}$

$\frac{d y}{d x}=-\frac{2}{1+x^2}$