Answer
$$ y'= -sech\, x-sech\, x \, csch^2x.$$
Work Step by Step
Since $ y=sech x\coth x $, then the derivative, by using the product rule $(uv)'=uv'+u'v $ and the facts that $(\ sech \ x)= -\ sech \ x \tanh x $ and $(\coth x)'=-csch^2 x $, is given by
$$ y'= -sech\, x\tanh x\coth x-sech\, x \, csch^2x\\
y'= -sech\, x-sech\, x \, csch^2x$$
