If $y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\sec ^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ then the value of $\frac{d y}{d x}$ at $x=\sqrt{3}$ is

$$ \cot ^{-1}\left(2 \cos \left(2 \operatorname{cosec}^{-1}(\sqrt{2})\right)\right)=\ldots $$

If $3 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)-4 \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\frac{\pi}{3}$ then the value of $x=$

If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}$ and $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$ then $2 x^2+y^2-x y=$ $\qquad$