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