Putting $x^2=\cos2z,1+x^2=1+\cos2z=2\cos^2z,1-x^2=1-\cos2z=2\sin^2z$
$\frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}=\frac{\cos z-\sin z}{\cos z+\sin z}=\frac{1-\tan z}{1+\tan z}=\tan\left(\frac\pi4-z\right)$
$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}=\tan^{-1}\tan\left(\frac\pi4-z\right)=\frac\pi4-z=\frac\pi4-\frac12\cos^{-1}x^2$
$\implies \frac{dy}{dx}=\frac{\frac\pi4-\frac12\cos^{-1}x^2}{dx}=-\frac12\left(-\frac{2x}{\sqrt{1-(x^2)^2}}\right)=\frac x{1-x^4}$