Answer
$y \cosh (xy-z^2)$ ;
$x \cosh (xy-z^2)$;
$-2z \cosh (xy-z^2)$
Work Step by Step
We need to take the first partial derivatives of the given function.
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa:
$f_x=\cosh (xy-z^2) \times (y) =y \cosh (xy-z^2)$
$f_y=\cosh (xy-z^2) \times (x) =x \cosh (xy-z^2)$
$f_z= \cosh (xy-z^2) \times (-2z) =-2z \cosh (xy-z^2)$
