[Example 5.3.4] Solve the system of equations:
$$
\left\{\begin{array}{l}
\left(1-x^{2}\right) y_{1}=2 x, \\
\left(1-y_{1}^{2}\right) y_{2}=2 y_{1}, \\
\left(1-y_{2}^{2}\right) y_{3}=2 y_{2}, \\
y_{3}=x .
\end{array}\right.
$$
This is a system of three ternary cubic equations involving $y_{1}, y_{2}, y_{3}$. Directly solving this system by substitution would make the equations increasingly complex, so the system needs to be transformed.

Observing the characteristics of the given equations, we can transform them as follows:
$$
y_{1}=\frac{2 x}{1-x^{2}},
$$

The right-hand side of the above equation resembles the double-angle formula for the tangent function. Therefore, we can set $x=\tan \theta$, which gives us:
$$
\begin{array}{c}
y_{1}=\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\tan 2 \theta, \\
y_{2}=\frac{2 y_{1}}{1-y_{1}^{2}}=\frac{2 \tan 2 \theta}{1-\tan ^{2} 2 \theta}=\tan 4 \theta, \\
y_{3}=\frac{2 y_{2}}{1-y_{2}^{2}}=\frac{2 \tan 4 \theta}{1-\tan ^{2} 4 \theta}=\tan 8 \theta .
\end{array}
$$

Given that $y_{3}=x=\tan \theta$, we have:
$$
\tan 8 \theta=\tan \theta .
$$

This is a trigonometric equation in $\theta$.
The work above essentially transforms the problem as follows:
$$
\begin{array}{l}
\text { The system of cubic equations } \\
\text { involving } y_{1}, y_{2}, y_{3} \\
\end{array} \Rightarrow \begin{array}{l}
\text { The trigonometric equation } \\
\text { involving } \theta: \tan 8 \theta=\tan \theta
\end{array}
$$

Now, we solve the equation:
$$
\begin{array}{c}
\tan 8 \theta=\tan \theta, \\
8 \theta=n \pi+\theta,(n \in \mathbf{Z}) \\
7 \theta=n \pi \\
\theta=\frac{n \pi}{7},(n=-3,-2,-1,0,1,2,3)
\end{array}
$$

Thus, the solutions to the original system are:
$$
\begin{aligned}
y_{1} & =\tan \frac{2 n \pi}{7}, \\
y_{2} & =\tan \frac{4 n \pi}{7}, \\
y_{3}=x=\tan \frac{n \pi}{7} \cdot(n & =-3,-2,-1,0,1,2,3)
\end{aligned}
$$