Solve the following system of equations:

$$
\begin{gathered}
\frac{x(z-1)}{y-1}=3 \\
\frac{x^{2}\left(z^{2}-1\right)}{y^{2}-1}=-3 \\
\frac{x^{3}\left(z^{3}-1\right)}{y^{3}-1}=9
\end{gathered}
$$
$$
\begin{gathered}
\frac{x(z-1)}{y-1}=3 \\
\frac{x^{2}\left(z^{2}-1\right)}{y^{2}-1}=-3 \\
\frac{x^{3}\left(z^{3}-1\right)}{y^{3}-1}=9
\end{gathered}
$$

Using (1) to transform (2) and (3):

$$
-3=\frac{x(z-1)}{y-1} \cdot \frac{x(z+1)}{y+1}=3 \cdot \frac{x(z+1)}{y+1}
$$

from which

$$
\frac{x(z+1)}{y+1}=-1
$$

Similarly, from (3) and (1):

$$
\frac{x^{2}\left(z^{2}+z+1\right)}{y^{2}+y+1}=3
$$

Under the conditions $y \neq 1$ and $y \neq -1$, (1) and (4) are equivalent to the equations

$$
\begin{aligned}
& x z-x=3 y-3 \\
& x z+x=-y-1
\end{aligned}
$$

from which

$$
x z=y-2, \quad \text { and } x=1-2 y .
$$

Substituting these values into the equation obtained by rearranging (5),

$$
(x z)^{2}+x(x z)+x^{2}=3 y^{2}+3 y+3
$$

after rearrangement, we get the equation $-6 y=0$. Thus, $y=0$. From (6), we can first determine the value of $x$, and then the value of $z$: $x=1, z=-2$. Since the value of $y$ obtained is different from 1 and -1, $x=1, y=0, z=-2$ is a solution to the system of equations, and there are no other solutions.