Let $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ and $z={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$

Putting $x=\tan \left(\theta \right),$ we have 

$\begin{array}{c}\Rightarrow y={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\left({\displaystyle \theta }\right)}-1}{\tan \left({\displaystyle \theta }\right)}\right)\\ y={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x\\ \therefore \frac{dy}{dx}=\frac{1}{2}\cdot \frac{1}{1+x^2}=\frac{1}{2}\text{ at }x=0\end{array}$

Again putting x = sin$\varphi$ , we have 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}={\tan }^{-1}\left(\frac{2\sin \varphi \cos \varphi }{1-2{\sin }^2\varphi }\right)={\tan }^{-1}\frac{\sin 2\varphi }{\cos 2\varphi }\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}={\tan }^{-1}\left(\tan 2\varphi \right)=2\varphi =2{\sin }^{-1}x\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dz}{dx}=\frac{2}{\sqrt{1-x^2}}=2\text{ at }x=0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dy}{dz}=\frac{dy}{dx}/\frac{dz}{dx}=\frac{1/2}{2}=\frac{1}{4}\end{array}$