Example 3.33. Find $\frac{\partial^{2} z}{\partial x^{2}}$ and $\frac{\partial^{2} z}{\partial y \partial x}$ of the implicit function from problem 3.32.
Solution. By the definition of the second-order partial derivative, we obtain

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

Now, substitute the expression for $z_{x}^{\prime}$ from problem (3.32):

$$
\begin{gathered}
\frac{\partial^{2} z}{\partial x^{2}}=\frac{\left(1+z^{2}\right) \frac{z}{z^{2}-1}+z\left(z^{2}-1\right)}{x^{2}\left(z^{2}-1\right)^{2}}=\frac{z\left(1+z^{2}+\left(z^{2}-1\right)^{2}\right)}{x^{2}\left(z^{2}-1\right)^{3}}= \\
=z \frac{z^{4}-z^{2}+2}{x^{2}\left(z^{2}-1\right)^{3}}
\end{gathered}
$$

Similarly, the mixed second-order partial derivative $\frac{\partial^{2} z}{\partial y \partial x}$ is found. We have

$$
\begin{aligned}
\frac{\partial^{2} z}{\partial y \partial x}= & \frac{\partial}{\partial y}\left(\frac{z}{x\left(z^{2}-1\right)}\right)=\frac{1}{x} \frac{z_{y}^{\prime}\left(z^{2}-1\right)-2 z^{2} z_{y}^{\prime}}{\left(z^{2}-1\right)^{2}}= \\
& =-\frac{z^{2}+1}{x\left(z^{2}-1\right)^{2}} z_{y}^{\prime}=-\frac{\left(z^{2}+1\right) z}{y x\left(z^{2}-1\right)^{3}}
\end{aligned}
$$

Above, we considered implicit functions defined by a single functional equation. In applications of analysis, it is often necessary to consider a set of $m \in \mathbb{N}$ implicit functions defined by a system of functional equations. We will consider the case of a set of two implicit functions $u=u(x, y)$ and $v=v(x, y)$, defined by a system of two functional equations

$$
\left\{\begin{array}{l}
F(x, y, u, v)=0 \\
G(x, y, u, v)=0
\end{array}\right.
$$

Let the system (3.22) have a solution $x=x_{0}, y=y_{0}, u=u_{0}$, $v=v_{0}$. Let the functions $F$ and $G$ have continuous first-order partial derivatives in a neighborhood of the point $M_{0}\left(x_{0}, y_{0}, u_{0}, v_{0}\right)$, and let the determinant

$$
J(u, v)=\frac{D(F, G)}{D(u, v)}=\left|\begin{array}{cc}
F_{u}^{\prime} & F_{v}^{\prime} \\
G_{u}^{\prime} & G_{v}^{\prime}
\end{array}\right|
$$

called the Jacobian, be non-zero at the point $M_{0}$. Then, in some neighborhood of the point $M_{0}$, the system defines a unique set of continuous functions $u=u(x, y)$ and $v=v(x, y)$ having continuous first-order partial derivatives, and such that $u\left(x_{0}, y_{0}\right)=u_{0}$ and $v\left(x_{0}, y_{0}\right)=v_{0}$.

To find the differentials $d u$ and $d v$, as well as $u_{x}^{\prime}, u_{y}^{\prime}, v_{x}^{\prime}, v_{y}^{\prime}$, we should write down the differentials $d F$ and $d G$:

$$
\begin{gathered}
d F=F_{x}^{\prime} d x+F_{y}^{\prime} d y+F_{u}^{\prime} d u+F_{v}^{\prime} d v=0 \\
d G=G_{x}^{\prime} d x+G_{y}^{\prime} d y+G_{u}^{\prime} d u+G_{v}^{\prime} d v=0
\end{gathered}
$$

and then solve the system of two equations (3.24) for $d u$ and $d v$.