Answer
$$ y'= -e^{\coth x}csch^2 (x) .$$
Work Step by Step
Recall that $(\coth x)'=-\csc^2 x$
Recall that $(e^x)'=e^x$
Since $ y=e^{\coth x}$, then the derivative, by using the chain rule, is given by
$$ y'=e^{\coth x} (\coth x)'= -e^{\coth x}csch^2 (x) .$$
