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

$$ \Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}$$

$$ \Rightarrow 2\,\cot \,y\, = u - {1 \over u}$$

where $$u = {x^x}$$

Differentiating both sides with respect to $$x,$$

we get $$ \Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}$$

$$ = \left( {1 + {1 \over {{u^2}}}} \right){{du} \over {dx}}$$

where $$u = {x^x} \Rightarrow \log \,u = x\,\log \,x$$

$$ \Rightarrow {1 \over u}{{du} \over {dx}} = 1 + \log \,x$$

$$ \Rightarrow {{du} \over {dx}} = {x^x}\left( {1 + \log \,x} \right)$$

$$\therefore$$ We get $$ - 2\cos e{c^2}y{{dy} \over {dx}}$$

$$ = \left( {1 + {x^{ - 2x}}} \right){x^x}\left( {1 + \log \,x} \right)$$

$$ \Rightarrow {{dy} \over {dx}} = {{\left( {{x^x} + {x^{ - x}}} \right)\left( {1 + \log x} \right)} \over { - 2\left( {1 + {{\cot }^2}y} \right)}}\,\,\,\,\,\,...\left( i \right)$$

Now when

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

gives $$1 - 2\,\cot y - 1 = 0$$

$$ \Rightarrow \,\,\cot y\, = 0$$

$$\therefore$$ From equation $$(i),$$ at $$x=1$$

and $$\cot \,y = 0,$$ we get

$$y'\left( 1 \right) = {{\left( {1 + 1} \right)\left( {1 + 0} \right)} \over { - 2\left( {1 + 0} \right)}} = - 1$$