It looks tough

$\begin{aligned} x^{2x}-2x^x\cot y -1& =0 \quad \quad \small \color{#3D99F6}{\text{When }x=1} \\ 1 - 2 \cot (y(1)) - 1 & = 0 \\ \implies \cot (y(1)) & = 0 \\ y(1) & = \frac{\pi}{2} \end{aligned}$

Now, we have:

$\begin{aligned} x^{2x}-2x^x\cot y -1& =0 \quad \quad \small \color{#3D99F6}{\text{Differentiate both sides with respect to }x} \\ 2(\ln x + 1)x^{2x} - 2(\ln x + 1)\cot y + 2x^x \csc^2 y \frac{dy}{dx} & = 0 \quad \quad \small \color{#3D99F6}{\text{When }x=1} \\ 2 - 0 + 2 \csc^2 \frac{\pi}{2} y'(1) & = 0 \\ \quad y'(1) & = \boxed{-1} \end{aligned}$

$\text{ We are Given } x^{2x}-2x^x\cot{y}-1=0 \\ (x^x)^2-2(x^x)(\cot{y})+\cot^2{y}-1-\cot^2{y}=0 \\ (x^x-\cot{y})^2=1+\cot^2{y} \\ x^x-\cot{y}=\csc{y} \\ x^x=\cot{y}+\csc{y} \\ \text{Taking Derivative w.r.t x both side} \\ x^x(1+\ln{x})=-\csc^2{y}\times y^{'}-\csc{y}\cot{y}\times y^{'} \\ x^x(1+\ln{x})=-\csc{y}(\csc{y}+\cot{y})y^{'} \\ y^{'}=-\dfrac{x^x(1+\ln{x})}{x^x\csc{y}} \\ y^{'}(1)=-\dfrac{1^1(1+\ln{1})}{\csc{(y(1))}} \implies -\sin{(y(1))} \\ \text{From our Original Equation , } 1^{2}-2 \times 1^1\cot{y(1)}-1=0 \implies y(1)=\dfrac{\pi}{2} \\ y^{'}(1)=-\sin{(y(1))} \implies \color{#3D99F6}{\boxed{-1}}$