Answer
$\dfrac{\partial z}{\partial x}=-\dfrac{2xy^2+yz\times sin(xyz)}{yxsin(xyz)+2z}$
and
$\dfrac{\partial z}{\partial y}=-\dfrac{2xy^2+xz\times sin(yxz)}{xysin(xyz)+2z}$
Work Step by Step
$cos(xyz)=1+x^2y^2+z^2$
Treat $z$ and $x$ as variables and $y$ as a constant.
$\dfrac{\partial [cos (xyz)]}{\partial x}=\dfrac{\partial (1)}{\partial x}+\dfrac{\partial (x^2y^2)}{\partial x}+\dfrac{\partial (z^2)}{\partial x}$
Use chain rule:
$-sin (xyz)\times \dfrac{\partial (xyz)}{\partial x}=2xy^2+2z\times \dfrac{\partial z}{\partial x}$
Thus, $\dfrac{\partial z}{\partial x}=-\dfrac{2xy^2+yz\times sin(xyz)}{yxsin(xyz)+2z}$
and
$\dfrac{\partial z}{\partial y}$ can be obtained from $\dfrac{\partial z}{\partial x}$ by replacing $x$ with $y$ and vice-versa.
$\dfrac{\partial z}{\partial y}=-\dfrac{2xy^2+xz\times sin(yxz)}{xysin(xyz)+2z}$
