Answer
$$ y'= \csc(9z+1)-9z\csc (9z+1) \cot (9z+1).$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\csc x)'=-\csc x \cot x$.
Since $ y=z\csc(9z+1)$, then by the product rule, the derivative $ y'$ is given by
$$ y'= \csc(9z+1)+9z(-\csc (9z+1) \cot (9z+1))\\= \csc(9z+1)-9z\csc (9z+1) \cot (9z+1).$$
