Answer
$\frac{\partial z}{\partial y}=-\frac{-x z-3 z^{2}}{-6 y z+xy}$
$\frac{\partial z}{\partial x}=-\frac{y z+2 x}{6 y z-x y}$
Work Step by Step
\[
0=-2+x y z+x^{2}-3 y z^{2}
\]
We have to find $\frac{\partial z}{\partial y}$ and $\frac{\partial z}{\partial x}$
\[
\begin{array}{l}
-\frac{\partial f / \partial x}{\partial f / \partial z} =\frac{\partial z}{\partial x} \\
-\frac{\partial f / \partial y}{\partial f / \partial z}=\frac{\partial z}{\partial y}
\end{array}
\]
We use this theorem in order to find it
\[
y z+2 x=\frac{\partial f}{\partial x}
\]
We find the partial derivative of $z, y$ and $x$
\[
\begin{array}{l}
-3 z^{2}+x z=\frac{\partial f}{\partial y} \\
-6 y z+x y=\frac{\partial f}{\partial z}
\end{array}
\]
We substitute
$\frac{\partial z}{\partial y}=-\frac{-x z-3 z^{2}}{-6 y z+xy}$
$\frac{\partial z}{\partial x}=-\frac{y z+2 x}{6 y z-x y}$
