Answer
$$ y'=\frac{(1+\tanh t) (-csch^2 t)-\coth t( \, sech^2 t)}{(1+\tanh t)^2} .$$
Work Step by Step
Since $ y=\frac{\coth t}{1+\tanh t} $, then the derivative $ y'$, using the quotient rule $(u/v)'=\frac{vu'-uv'}{v^2}$ and the facts that $(\tanh t)'=sech^2t $ $(\coth t)'=-csch^2t $, is given by $$ y'=\frac{(1+\tanh t) (-csch^2 t)-\coth t( \, sech^2 t)}{(1+\tanh t)^2} .$$
