Given that

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

Put $x^2 = \cos 2\theta$

$\Rightarrow \theta =\frac{1}{2}{\cos }^{-1}{x}^2 --\left(I\right)$

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

We know that $\cos 2\theta = 1 - 2{\sin }^2\theta = 2{\cos }^2\theta - 1$

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

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

$y = {\tan }^{-1}\left(\tan \left(\frac{\pi }{4}+\theta \right)\right)$

$\Rightarrow y = \frac{\pi }{4} + \theta$

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

Differentiate with respect to $x$ on both sides

$\Rightarrow \frac{dy}{dx} = \frac{1}{2}\frac{-1}{\sqrt{1-x^4}}\times 2x$

$\Rightarrow \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^4}}$