$f\left(x\right)={\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

Let, $x=\cos 2\theta$

$\Rightarrow 1+x=1+\cos 2\theta$

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

$\Rightarrow \sqrt{1+x}=\sqrt{2}\cos \theta$

Also, $1-x=1-\cos 2\theta$

$\Rightarrow 1-x=2{\sin }^2\theta$

$\Rightarrow \sqrt{1-x} =\sqrt{2}\sin \theta$

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

$\Rightarrow f\left(x\right)={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right)$

$\Rightarrow f\left(x\right)={\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{4}-\theta \right)\right)$

$\Rightarrow f\left(x\right)=\frac{\mathrm{\pi }}{4}-\theta =\frac{\mathrm{\pi }}{4}-\frac{{\cos }^{-1}x}{2}$

We need to find, $\underset{x\to \frac{1}{2}}{\lim }2\left[\frac{f\left(x\right)-f\left({\displaystyle \frac{1}{2}}\right)}{2x-1}\right]$

Applying L Hospital's rule

$=\underset{x\to \frac{1}{2}}{\lim }\left[f\text{'}\left(x\right)\right]$

$=f\text{'}\left(\frac{1}{2}\right)$

Now, $f\left(x\right)=\frac{1}{2}\left[\frac{\mathrm{\pi }}{2}-{\cos }^{-1}x\right]$

$\Rightarrow f\text{'}\left(x\right)=\frac{1}{2}\left[0-\frac{-1}{\sqrt{1-x^2}}\right]$

$\Rightarrow f\text{'}\left(x\right)=\frac{1}{2}\left[\frac{1}{\sqrt{1-x^2}}\right]$

$\Rightarrow f\text{'}\left(\frac{1}{2}\right)=\frac{1}{2}\left[\frac{1}{\sqrt{1-{\left({\displaystyle \frac{1}{2}}\right)}^2}}\right]$

$\Rightarrow f\text{'}\left(\frac{1}{2}\right)=\frac{1}{\sqrt{3}}$