Example 1 Given $x_{n+1}=\frac{x_{n}\left(x_{n}^{2}+3 a^{2}\right)}{3 x_{n}^{2}+a^{2}}, x_{1}=$ $b$, where $a, b$ are real constants, $a \neq b$. Find $x_{n}$.
Solution: Let $\alpha$ replace $x_{n+1}$ and $x_{n}$, we get the fixed points $\alpha=0$, $\pm a$. Therefore,
$$
\begin{array}{l}
\frac{x_{n+1}-a}{x_{n+1}+a}=\frac{\frac{x_{n}\left(x_{n}^{2}+3 a^{2}\right)}{3 x_{n}^{2}+a^{2}}-a}{\frac{x_{n}\left(x_{n}^{2}+3 a^{2}\right)}{3 x_{n}^{2}+a}+a} \\
=\frac{\left(x_{n}-a\right)^{3}}{\left(x_{n}+a\right)^{3}} . \\
\therefore \frac{x_{n}-a}{x_{n}+a}=\left(\frac{x_{1}-a}{x_{1}+a}\right)^{3^{n-1}}=\frac{(b-a)^{3^{n-1}}}{(b+a)^{3^{n-1}}} .
\end{array}
$$

Solving for $x_{n}$, we get $x_{n}=\frac{a\left[(b+a)^{3^{n-1}}+(b-a)^{3^{n-1}}\right]}{\left[(b+a)^{3^{n-1}}-(b-a)^{3^{n-1}}\right]}$.
Similarly, we can solve
$x_{n+1}=\frac{x_{n}^{2}+l}{2 x_{n}^{2}+k}$. $(k, l$ are real constants)