If ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\pi }{3},x>0$  then $\text{'}x\text{'}=$

${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$

$2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}\Rightarrow {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{6}$

$\frac{2x}{1-x^2}=\frac{1}{\sqrt{3}}\Rightarrow 2\sqrt{3}x=1-x^2\Rightarrow x^2+2\sqrt{3}x-1=0$

$x=\frac{-2\sqrt{3}\pm \sqrt{{\left(2\sqrt{3}\right)}^2-4\left(-1\right)}}{2}=\frac{-2\sqrt{3}\pm \sqrt{16}}{2}$

$x=\frac{-2\sqrt{3}\pm 4}{2}=-\sqrt{3}\pm 2$