Given equation

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

We need to find $y'(1)$.

First find $y$ at $x=1$

Substitute $x=1$:

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

Hence,

$y=\frac{\pi}{2}+n\pi$

Therefore,

$\csc^2 y=1$

Differentiate implicitly

The equation is

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

Differentiate both sides w.r.t. $x$.

Recall:

$\frac{d}{dx}(x^x)=x^x(\ln x+1)$

Also,

$x^{2x}=e^{2x\ln x} \implies \frac{d}{dx}(x^{2x})=x^{2x}(2\ln x+2)$

Now differentiate term by term:

$\frac{d}{dx}(x^{2x})-2\frac{d}{dx}(x^x\cot y)=0$

Using product rule on $x^x\cot y$:

$\frac{d}{dx}(x^x\cot y)=x^x(\ln x+1)\cot y+x^x\frac{d}{dx}(\cot y)$

and

$\frac{d}{dx}(\cot y)=-\csc^2 y\, y'$

So,

$x^{2x}(2\ln x+2)-2\left[x^x(\ln x+1)\cot y-x^x\csc^2 y\, y'\right]=0$

Substitute $x=1$

At $x=1$,

$x^{2x}=1, \quad x^x=1, \quad \ln 1=0, \quad \cot y=0, \quad \csc^2 y=1$

Therefore,

$1\cdot (2)-2\left[1\cdot (1)\cdot 0-1\cdot 1\cdot y'\right]=0$

$2-2( - y')=0$

$2+2y'=0$

$y'=-1$

Final answer

$y'(1)=-1$

So the correct option is D.