Let $f\left(x\right)=\cos \left(2{\tan }^{-1}\sin \left({\cot }^{-1}\sqrt{\frac{1-x}{x}}\right)\right),0<x<1$. Then:

Put $x={\sin }^2\theta ,0<x<1$

$\sin \theta =\sqrt{x}$

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

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

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

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

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

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{\left(1+x\right)\left(-1\right)-\left(1-x\right)·1}{{\left(1+x\right)}^2}$

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

Multiply, ${\left(1-x\right)}^2$ on both sides

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

Now, $2{\left(f\left(x\right)\right)}^2=\frac{2{\left(1-x\right)}^2}{{\left(1+x\right)}^2}$

$\therefore {\left(1-x\right)}^2{f}^{\text{'}}\left(x\right)+2{\left(f\left(x\right)\right)}^2=0$ option $\left(1\right)$ satisfied