Solution. We know that f(x) = \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{e^x + e^{-x}}{e^x - e^{-x}}. So there are two different ways to prove this. The first way is to do it via f(x) = \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{g(x)}{h(x)}. We will apply the quotient rule here:
\begin{align*}
f'(x) = \frac{d}{dx} \coth(x) = \frac{\cosh(x)}{\sinh(x)} = \frac{g(x)}{h(x)} = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}.
\end{align*}\begin{align*}
f'(x) &= \frac{d}{dx} \coth(x) \\
&= \frac{\cosh(x)}{\sinh(x)} \\
&= \frac{\sinh(x)\sinh(x) - \cosh(x)\cosh(x)}{\sinh^2(x)} \\
&= \frac{\sinh^2(x) - \cosh^2(x)}{\sinh^2(x)} \\
&= \frac{-1}{\sinh^2(x)} \\
&= -\text{csch}^2(x).
\end{align*}