Given,
$$
\begin{gathered}
f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right) \\
y=\cot ^{-1}\left(\frac{x^{2 x}-1}{2 \cdot x^x}\right)
\end{gathered}
$$

Put $x^x=\tan \theta$
$$
\begin{aligned}
\therefore \quad y & =\cot ^{-1}\left(\frac{\tan ^2 \theta-1}{2 \tan \theta}\right) \\
& =\cot ^{-1}(-\cot 2 \theta) \\
y & =\pi-\cot ^{-1}(\cot 2 \theta) \\
y & =\pi-2 \theta \\
y & =\pi-2 \tan ^{-1}\left(x^x\right) \\
\therefore \quad \frac{d y}{d x} & =-\frac{2}{1+x^{2 x}} \cdot x^x(1+\log x)
\end{aligned}
$$

So, $\left.\quad \frac{d y}{d x}\right|_{x=1}=\frac{-2}{1+(1)^2} \cdot 1^1(1+\log 1)$
$$
=-\frac{2}{2} \cdot 1=-1
$$