Example 2. Find the total differential of the function $z=z(x, y)$, given by the equation $e^{x y z}-\operatorname{arctg} \frac{x y}{z}=0$.
Solution. In this case, $F(x, y, z)=e^{x y z}-\operatorname{arctg} \frac{x y}{z}$. First, we find the partial derivatives of the function $F(x, y, z)$:

$$
\begin{gathered}
F_{x}^{\prime}=\frac{\partial F}{\partial x}=y z e^{x y z}-\frac{1}{1+\frac{x^{2} y^{2}}{z^{2}}} \cdot \frac{y}{z}=y z\left(e^{x y z}-\frac{1}{x^{2} y^{2}+z^{2}}\right) \\
F_{y}^{\prime}=\frac{\partial F}{\partial y}=x z e^{x y z}-\frac{1}{1+\frac{x^{2} y^{2}}{z^{2}}} \cdot \frac{x}{z}=x z\left(e^{x y z}-\frac{1}{x^{2} y^{2}+z^{2}}\right) \\
F_{z}^{\prime}=\frac{\partial F}{\partial z}=x y e^{x y z}-\frac{1}{1+\frac{x^{2} y^{2}}{z^{2}}} \cdot\left(\frac{-x y}{z^{2}}\right)=x y\left(e^{x y z}+\frac{1}{x^{2} y^{2}+z^{2}}\right)
\end{gathered}
$$

Using formulas (7.17), we obtain the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$:

$$
\begin{gathered}
\frac{\partial z}{\partial x}=-\frac{y z\left(e^{x y z}-\frac{1}{x^{2} y^{2}+z^{2}}\right)}{x y\left(e^{x y z}+\frac{1}{x^{2} y^{2}+z^{2}}\right)}=\frac{z\left[1-\left(x^{2} y^{2}+z^{2}\right) e^{x y z}\right]}{x\left[1+\left(x^{2} y^{2}+z^{2}\right) e^{x y z}\right]} \\
\frac{\partial z}{\partial y}=-\frac{x z\left(e^{x y z}-\frac{1}{x^{2} y^{2}+z^{2}}\right)}{x y\left(e^{x y z}+\frac{1}{x^{2} y^{2}+z^{2}}\right)}=\frac{z\left[1-\left(x^{2} y^{2}+z^{2}\right) e^{x y z}\right]}{y\left[1+\left(x^{2} y^{2}+z^{2}\right) e^{x y z}\right]}
\end{gathered}
$$

## Therefore,

$$
d z=\frac{\partial z}{\partial x} d x+\frac{\partial z}{\partial y} d y=\frac{1-\left(x^{2} y^{2}+z^{2}\right) e^{x y z}}{1+\left(x^{2} y^{2}+z^{2}\right) e^{x y z}} \cdot \frac{z}{x y}(y d x+x d y)
$$