Let y be an implicit function of x defined by $x^{2x}-2x^x\cot y-1=0,$ then $y^1\left(1\right)$ is equal to 

$x^{2x}-2x^x\cot y-1=0$

$\text{ Divide with }{x}^x\Rightarrow \cot y=\frac{x^x-x^{-x}}{2}$

$\Rightarrow y={\cot }^{-1}\left(\frac{x^x-x^{-x}}{2}\right);$

$\frac{dy}{dx}=\frac{-1}{1+{\left(\frac{x^x-x^{-x}}{2}\right)}^2}\frac{d}{dx}\left(\frac{x^x-x^{-x}}{2}\right)$

$=-\frac{4}{x^{2x}+x^{-2x}+2}\left[\frac{1}{2}\left(x^x+x^{-x}\right)(1+\log x)\right]$

$\Rightarrow y^{\text{'}}\left(1\right)=-\frac{2}{1+1+2}\left[2\right(1+\log 1\left)\right]=-1$