Answer
$\frac{\partial z}{\partial y}=-\frac{x e^{x y} \cos (y z)-z e^{x y} \sin (y z)-z e^{y z} \sin (x z)}{-y e^{x y} \sin (y z)-x e^{y z} \cos (x z)-y e^{y z} \sin (x z)}$
$\frac{\partial z}{\partial x}=-\frac{-z e^{y z} \cos (x z)+y e^{x y} \cos (y z)}{-y e^{x y} \sin (y z)-x e^{y z} \cos (x z)-y e^{y z} \sin (x z)}$
Work Step by Step
We have to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$
$0=2+e^{x y} \cos (y z)-e^{y z} \sin (x z)$
We use this theorem in order to find it
$\frac{\partial z}{\partial x}=-\frac{\partial f / \partial x}{\partial f / \partial z}$
$\frac{\partial z}{\partial y}=-\frac{\partial f / \partial y}{\partial f / \partial z}$
We find partial derivates of $z, y$ and $x$
$-z e^{y z} \cos (x z)+y e^{x y} \cos (y z)=\frac{\partial f}{\partial x}$
$x e^{x y} \cos (y z)-z e^{x y} \sin (y z)-z e^{y z} \sin (x z)=\frac{\partial f}{\partial y}$
$-x e^{y z} \cos (x z)-y e^{y z} \sin (x z)-y e^{x y} \sin (y z)=\frac{\partial f}{\partial z}$
We substitute
$\frac{\partial z}{\partial y}=-\frac{x e^{x y} \cos (y z)-z e^{x y} \sin (y z)-z e^{y z} \sin (x z)}{-y e^{x y} \sin (y z)-x e^{y z} \cos (x z)-y e^{y z} \sin (x z)}$
$\frac{\partial z}{\partial x}=-\frac{-z e^{y z} \cos (x z)+y e^{x y} \cos (y z)}{-y e^{x y} \sin (y z)-x e^{y z} \cos (x z)-y e^{y z} \sin (x z)}$
