Answer
$$\frac{{dy}}{{dx}} = \frac{{1 + x\coth x}}{{\operatorname{csch} x}}$$
Work Step by Step
$$\eqalign{
& y = x/\operatorname{csch} x \cr
& {\text{computing }}dy/dx \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {x/\operatorname{csch} x} \right) \cr
& {\text{by the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\operatorname{csch} x\left( x \right)' - x\left( {\operatorname{csch} x} \right)'}}{{{{\left( {\operatorname{csch} x} \right)}^2}}} \cr
& {\text{using basic formulas for differentiation}} \cr
& \frac{{dy}}{{dx}} = \frac{{\operatorname{csch} x\left( 1 \right) - x\left( { - \operatorname{csch} x\coth x} \right)}}{{{{\left( {\operatorname{csch} x} \right)}^2}}} \cr
& {\text{multiplying}} \cr
& \frac{{dy}}{{dx}} = \frac{{\operatorname{csch} x + xcschx\coth x}}{{{{\operatorname{csch} }^2}x}} \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{1 + x\coth x}}{{\operatorname{csch} x}} \cr} $$
