Answer
$$y' = \frac{1}{{{x^2}}}\operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right)$$
Work Step by Step
$$\eqalign{
& y = \operatorname{csch} \left( {1/x} \right) \cr
& {\text{find the derivatvive}} \cr
& y' = \left[ {\operatorname{csch} \left( {\frac{1}{x}} \right)} \right]' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{ see page 476}}} \right) \cr
& \frac{d}{{dx}}\left[ {\operatorname{csch} \left( {\frac{1}{x}} \right)} \right] = - \operatorname{csch} u\coth u\frac{{du}}{{dx}} \cr
& y' = - \operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right)\left( {\frac{1}{x}} \right)' \cr
& {\text{compute derivative}} \cr
& y' = - \operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr
& {\text{simplifying}} \cr
& y' = \frac{1}{{{x^2}}}\operatorname{csch} \left( {\frac{1}{x}} \right)\coth \left( {\frac{1}{x}} \right) \cr} $$
