Answer
$$
x^{2}+2y^{2}+3z^{2}=1
$$
To find $\frac{\partial z}{\partial x}$, we differentiate implicitly with respect to x, being careful to treat $y$ as a constant:
$$
\begin{aligned}
\frac{\partial z }{\partial x}&= \frac{-x}{3z }.
\end{aligned}
$$
To find $\frac{\partial z}{\partial y}$, we differentiate implicitly with respect to x, being careful to treat $x$ as a constant:
$$
\begin{aligned}
\frac{\partial z }{\partial y}&= \frac{-2y}{3z }.
\end{aligned}
$$
Work Step by Step
$$
x^{2}+2y^{2}+3z^{2}=1
$$
To find $\frac{\partial z}{\partial x}$, we differentiate implicitly with respect to x, being careful to treat $y$ as a constant:
$$
\begin{aligned}
\frac{\partial }{\partial x} (x^{2}+2y^{2}+3z^{2}) &=\frac{\partial }{\partial x} (1)\\
\frac{\partial }{\partial x} (x^{2})+2\frac{\partial }{\partial x} ( y^{2}) +3 \frac{\partial }{\partial x} (z^{2} )&=\frac{\partial }{\partial x} (1) \\
2x +2(0)+3(2z \frac{\partial z }{\partial x}))&=(0)
\\
2x +6z \frac{\partial z }{\partial x}&=0
\\
6z \frac{\partial z }{\partial x}&=-2x
\\
\frac{\partial z }{\partial x}&= \frac{-2x}{6z }= \frac{-x}{3z }.
\end{aligned}
$$
And to find $\frac{\partial z}{\partial y}$, we differentiate implicitly with respect to x, being careful to treat $x$ as a constant:
$$
\begin{aligned}
\frac{\partial }{\partial y} (x^{2}+2y^{2}+3z^{2}) &=\frac{\partial }{\partial y} (1)\\
\frac{\partial }{\partial y} (x^{2})+2\frac{\partial }{\partial y} ( y^{2}) +3 \frac{\partial }{\partial y} (z^{2} )&=\frac{\partial }{\partial y} (1) \\
(0) +2(2y)+3(2z \frac{\partial z }{\partial y}))&=(0)
\\
4y +6z \frac{\partial z }{\partial y}&=0
\\
6z \frac{\partial z }{\partial y}&=-4y
\\
\frac{\partial z }{\partial y}&= \frac{-4y}{6z }= \frac{-2y}{3z }.
\end{aligned}
$$
