Example 3.32. Find the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ of the function $z=f(x, y)$, given implicitly by the equation $e^{z^{2}}-$ $-x^{2} y^{2} z^{2}=0$
Solution. The partial derivatives $z_{x}^{\prime}$ and $z_{y}^{\prime}$ are found in this case using formulas (3.20). We need to find the necessary partial derivatives of the function $F(x, y, z)$, which is defined, obviously, by the expression

$$
F(x, y, z)=e^{z^{2}}-x^{2} y^{2} z^{2}
$$

## We have

$$
F_{x}^{\prime}=-2 x y^{2} z^{2} ; F_{y}^{\prime}=-2 x^{2} y z^{2} ; F_{z}^{\prime}=2 z e^{z^{2}}-2 x^{2} y^{2} z
$$

Now, according to (3.20), we get

$$
\frac{\partial z}{\partial x}=-\frac{-2 x y^{2} z^{2}}{2 z\left(e^{z^{2}}-x^{2} y^{2}\right)} ; \frac{\partial z}{\partial y}=-\frac{-2 x^{2} y z^{2}}{2 z\left(e^{z^{2}}-x^{2} y^{2}\right)}
$$

In the obtained expressions for the partial derivatives $z_{x}^{\prime}$ and $z_{y}^{\prime}$, we can (but do not have to) replace $e^{z^{2}}$ with $x^{2} y^{2} z^{2}$ according to the equation defining the implicit function. As a result, we get

$$
\begin{aligned}
\frac{\partial z}{\partial x} & =\frac{x y^{2} z^{2}}{z\left(x^{2} y^{2} z^{2}-x^{2} y^{2}\right)}=\frac{z}{x\left(z^{2}-1\right)} \\
\frac{\partial z}{\partial y} & =\frac{x^{2} y z^{2}}{x^{2} y^{2} z^{2}-x^{2} y^{2}}=\frac{z}{y\left(z^{2}-1\right)}
\end{aligned}
$$