Let \(y= \operatorname{csch} (x)\):
\[\begin{gathered} y = \operatorname{csch}(x) \\ \operatorname{arcsch}(y) = x \end{gathered}\]
This means:
\[ \begin{gathered} \frac{dy}{dx} = - \operatorname{csch}(x) \coth(x) \\ \frac{d}{dy} \operatorname{arcsch}(y) = \frac{dx}{dy} = \frac{1}{- \operatorname{csch}(x) \coth(x) }\end{gathered} \]
Since \(\operatorname{csch}^2(x) = \coth^2(x) -1\):
\[ \frac{dx}{dy} = \frac{-1}{\operatorname{csch}(x) \left( ±\sqrt{\operatorname{csch}^2(x)+1} \right)} \]
Keep in mind the sign of \(\operatorname{csch}\) and \(\coth(x)\) are the same. If \(\coth(x)\) is negative then \(\operatorname{csch}(x)\) is negative, and vice versa. If \(\coth(x)\) is positive then \(\operatorname{csch}(x)\) is positive, and vice versa.
\[\begin{align} \frac{dx}{dy} &= \frac{-1}{\operatorname{csch}(x) \frac{|\operatorname{csch}(x)|}{\operatorname{csch}(x)} |\sqrt{\operatorname{csch}^2(x)+1}| } \\ &= \frac{-1}{y \frac{|y|}{y} \sqrt{y^2+1}} = \frac{-1}{|y| |\sqrt{y^2+1}|} \end{align}\]