Answer
$$
z_x= 2xy\cosh (x^2y),\quad
z_y= x^2\cosh (x^2y).
$$
Work Step by Step
Recall that $(\sinh x)'=\cosh x$
Recall that $(x^n)'=nx^{n-1}$
Since $ z=\sinh (x^2y)$, by using the chain rule, we have
$$
z_x= \cosh(x^2y) (2xy)=2xy\cosh (x^2y),\\
z_y= \cosh(x^2y) (x^2)=x^2\cosh (x^2y).
$$
