## Calculus 8th Edition

$\dfrac{\partial z}{\partial x}=-\dfrac{2xy^2+yz\times sin(xyz)}{yxsin(xyz)+2z}$ and $\dfrac{\partial z}{\partial y}=-\dfrac{2xy^2+xz\times sin(yxz)}{xysin(xyz)+2z}$
$cos(xyz)=1+x^2y^2+z^2$ Treat $z$ and $x$ as variables and $y$ as 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 by from $\dfrac{\partial z}{\partial x}$ by replacing $x$ with $y$ and vice-versa. $\dfrac{\partial z}{\partial y}=-\dfrac{2xy^2+xz\times sin(yxz)}{xysin(xyz)+2z}$