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