Answer
$\frac{\partial z}{\partial y}=-\frac{z^{2}x \cos (x y z)}{\sin (x y z)+x y z \cos (x y z)}$
$\frac{\partial z}{\partial x}=-\frac{y z^{2} \cos (x y z)+2 x}{\sin (x y z)+x y z \cos (x y z)}$
Work Step by Step
We are given that
$0=z \sin (x y z)+x^{2}$
Differentiating
$\frac{\partial}{\partial x}\left(z \sin (x y z)+x^{2}\right)=\frac{d}{d x}(0)$
$0=\sin (x y z) \frac{\partial z}{\partial x}+2 x+z \cos (x y z) *\left(y z+x y \frac{\partial z}{\partial x}\right)$
$(\sin (x y z)+x y z \cos (x y z)) \frac{\partial z}{2}$
$\frac{\partial z}{\partial x}=-\frac{y z^{2} \cos (x y z)+2 x}{\sin (x y z)+x z z \cos (x z z)}$
Differentiating
$\frac{\partial}{\partial y}\left(x^{2}+z \sin (x y z)\right)=\frac{d}{d y}(0)$
$\sin (x y z) \frac{\partial z}{\partial y}+z \cos (x y z) *\left(x z+x y \frac{\partial z}{\partial y}\right)=0$
$\frac{\partial z}{\partial y}{\sin (x y z)+x y z \cos (x y z)}=-{z^{2}x \cos (x y z)}$
$\frac{\partial z}{\partial y}=-\frac{z^{2}x \cos (x y z)}{\sin (x y z)+x y z \cos (x y z)}$
