Answer
$$\frac{\partial z }{\partial x}=-\frac{x}{z}
$$
$$ \frac{\partial z }{\partial y}=-\frac{y}{z}$$
Work Step by Step
Given $$ x^{2}+y^{2} +z^2=1$$
$$ \Rightarrow x^{2}+y^{2} +z^2-1=0$$
by letting $$F(x,y,z)= x^{2}+y^{2} +z^2-1$$
So, we have
$$F_x(x,y,z)=\frac{\partial F(x,y,z)}{\partial x}=2x\\
$$
$$F_y(x,y,z)=\frac{\partial F(x,y,z)}{\partial y}= 2y$$
$$F_z(x,y,z)=\frac{\partial F(x,y,z)}{\partial z}= 2z$$
Also, we get
\begin{aligned} \frac{\partial z }{\partial x}=&-\frac{F_{x}(x, y,z)}{F_{z}(x, y,z)} \\
&=- \frac{2x}{2z} \\ &=-\frac{x}{z}\end{aligned}
\begin{aligned} \frac{\partial z }{\partial y}=&-\frac{F_{y}(x, y,z)}{F_{z}(x, y,z)} \\
&=- \frac{2y}{2z} \\ &=-\frac{y}{z}\end{aligned}
