Answer
\begin{aligned} \frac{\partial z }{\partial x} =-\frac{\ln y}{y^2+2z}
\end{aligned}
\begin{aligned}
\frac{\partial z }{\partial y} =-\frac{x+2y^2z}{y^3+2yz} \\
\end{aligned}
Work Step by Step
Given $$x \ln y+y^{2} z+z^{2}=8$$
$$\Rightarrow x \ln y+y^{2} z+z^{2}-8=0$$
by letting $$F(x,y,z)= x \ln y+y^{2} z+z^{2}-8$$
So, we have
$$F_x(x,y,z)=\frac{\partial F(x,y,z)}{\partial x}=\ln y\\
$$
$$F_y(x,y,z)=\frac{\partial F(x,y,z)}{\partial y}= \frac{x}{y}+2yz$$
$$F_z(x,y,z)=\frac{\partial F(x,y,z)}{\partial z}=y^2+2z$$
Also, we get
\begin{aligned} \frac{\partial z }{\partial x}&=-\frac{F_{x}(x, y,z)}{F_{z}(x, y,z)} \\
&=-\frac{\ln y}{y^2+2z}
\end{aligned}
\begin{aligned}
\frac{\partial z }{\partial y}&=-\frac{F_{y}(x, y,z)}{F_{z}(x, y,z)} \\
&=-\frac{\frac{x}{y}+2yz}{y^2+2z} \\
&=-\frac{x+2y^2z}{y^3+2yz} \\
\end{aligned}
