Let \(y= \coth (x)\):
\[\begin{gathered} y = \coth(x) \\ \operatorname{arcoth}(y) = x \end{gathered}\]
This means:
\[\begin{gathered} \frac{dy}{dx} = - \operatorname{csch}^2(x) \\ \frac{d}{dy} \operatorname{arcoth}(y) = \frac{dx}{dy} = \frac{-1}{\operatorname{csch}^2(x)} \end{gathered}\]
Since \(\operatorname{csch}^2(x) = \coth^2(x) - 1\):
\[\begin{align} \frac{dx}{dy} &= \frac{-1}{\coth^2(x)-1} \\ &= \frac{1}{1-y^2} \end{align}\]
Since \(y= \coth (x)\), then \(|y| \gt 1 \).