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