Answer
$$y \cosh (xy-z^2) \\ x \cosh (xy-z^2) \\-2z \cosh (xy-z^2)$$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$ , by treating $y$ and $z$ as a constant.
$f_x=y \times \cosh (xy-z^2) =y \cosh (xy-z^2)$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $y$, by treating $x$ and $z$ as a constant.
$f_y=x \times\cosh (xy-z^2) =x \cosh (xy-z^2)$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $z$, by treating $x$ and $y$ as a constant.
$f_z= (-2z) \times\cosh (xy-z^2) =-2z \cosh (xy-z^2)$