Example 7 Given an infinite sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{0}=x, a_{1}=y, \\
a_{n+1}=\frac{a_{n} a_{n-1}+1}{a_{n}+a_{n-1}}(n=1,2, \ldots) .
\end{array}
$$
(1) For which real numbers $x, y$, does there always exist a positive integer $n_{0}$, such that when $n \geqslant n_{0}$, $a_{n}$ is always a constant?
(2) Find the general term formula of the sequence $\left\{a_{n}\right\}$.
(2006, National High School Mathematics Competition)
Analysis: This is a fractional recursive sequence, and the fixed point method easily comes to mind. By transformation, it can be converted to $b_{n+1} = b_{n} b_{n-1} (n \in \mathbf{N}_{+})$. During the iteration, it can be observed that the exponents are related to the Fibonacci sequence, which serves as a starting point to transform the non-linear problem into a linear one.

Solution: From the recursive equation $f(x) = \frac{x^2 + 1}{2x} = x$, we get the fixed points $x = \pm 1$. Using the fixed point method,
$$
\begin{array}{l}
\frac{a_{n+1} - 1}{a_{n+1} + 1} = \frac{\frac{a_{n} a_{n-1} + 1}{a_{n} + a_{n-1}} - 1}{\frac{a_{n} a_{n-1} + 1}{a_{n} + a_{n-1}} + 1} \\
= \frac{a_{n} a_{n-1} + 1 - a_{n} - a_{n-1}}{a_{n} a_{n-1} + 1 + a_{n} + a_{n-1}} \\
= \frac{(a_{n} - 1)(a_{n-1} - 1)}{(a_{n} + 1)(a_{n-1} + 1)} \\
= \frac{a_{n-1} - 1}{a_{n} + 1} \cdot \frac{a_{n-1} - 1}{a_{n-1} + 1}. \\
\text{Let } b_{n} = \frac{a_{n} - 1}{a_{n} + 1}, \text{ then} \\
b_{n+1} = b_{n} b_{n-1} (n \in \mathbf{N}_{+}).
\end{array}
$$

It is clear that $b_{0} = \frac{x - 1}{x + 1}, b_{1} = \frac{y - 1}{y + 1}$.
Notice that
$$
\begin{array}{l}
b_{n} = b_{n-1} b_{n-2} = (b_{n-2} b_{n-3}) b_{n-2} \\
= b_{n-2}^2 b_{n-3} = b_{n-3}^3 b_{n-4}^2 = \cdots \\
= b_{1}^{F} b_{n-1}^{F} b_{0}^{F_{n-2}},
\end{array}
$$

where $F_{n+1} = F_{n} + F_{n-1}, F_{0} = F_{1} = 1, \{F_{n}\}$ is the Fibonacci sequence.
$$
\begin{array}{l}
\text{Thus, } b_{n} = \frac{a_{n} - 1}{a_{n} + 1} = b_{1}^{F_{1-1}-1} b_{0}^{F_{n-2}} \\
= \left(\frac{y - 1}{y + 1}\right)^{F_{n-1}} \left(\frac{x - 1}{x + 1}\right)^{F_{n-2}}. \\
\text{Hence, } \frac{a_{n} - 1}{a_{n} + 1} = \left(\frac{y - 1}{y + 1}\right)^{F_{n-1}} \left(\frac{x - 1}{x + 1}\right)^{F_{n-2}} (n \geqslant 2).
\end{array}
$$
(1) To ensure that there exists a positive integer $n_{0}$ such that when $n \geqslant n_{0}$, $a_{n}$ is always a constant, we need to consider different cases.
(i) If $n_{0} = 1$, when $n \geqslant n_{0}$, $a_{n}$ is always a constant.
By $a_{1} = y$,
$$
\begin{array}{l}
a_{2} = \frac{a_{0} a_{1} + 1}{a_{0} + a_{1}} = \frac{x y + 1}{x + y} = y, \\
a_{3} = \frac{y^2 + 1}{2 y} = y,
\end{array}
$$

we have $|y| = 1$, and $y \neq -x$.
In this case, $a_{n}$ is always the constant 1 or -1.
(ii) If $n_{0} \geqslant 2$ when $n \geqslant n_{0}$, $a_{n}$ is always a constant.

First, when $a_{n} = -1 (n \geqslant n_{0})$, if $n_{0} \geqslant 2$, by $a_{n_{0}} = -1, a_{n_{0}+1} = -1$ and $a_{n_{0}+1} = \frac{a_{n_{0}} a_{n_{0}-1} + 1}{a_{n_{0}} + a_{n_{0}-1}}$, we have $a_{n_{0}-1} \neq 1$.
Notice that $a_{n_{0}-1} \neq -1$.
Also, by $a_{n_{0}} = \frac{a_{n_{0}-1} a_{n_{0}-2} + 1}{a_{n_{0}-1} + a_{n_{0}-2}}$, we have $a_{n_{0}-2} = -1$.
Thus, by $a_{n_{0}-1} = \frac{a_{n_{0}-2} a_{n_{0}-3} + 1}{a_{n_{0}-2} + a_{n_{0}-3}}$, we have $a_{n_{0}-1} = -1$, which is a contradiction.

Therefore, it can only be $n_{0} = 2$, i.e., $a_{n} = -1 (n \geqslant 2)$, so
$$
\begin{array}{l}
a_{2} = \frac{a_{0} a_{1} + 1}{a_{0} + a_{1}} = \frac{x y + 1}{x + y} = -1, \\
a_{3} = \frac{a_{2} a_{1} + 1}{a_{2} + a_{1}} = \frac{-1 \cdot y + 1}{-1 + y} = -1, \\
\ldots \ldots \\
\text{Thus, } \frac{x y + 1}{x + y} = -1, \text{ and } y \neq 1 \\
\Rightarrow x y + x + y + 1 = 0, \text{ and } y \neq -x, y \neq 1 \\
\Rightarrow x = -1 \text{ or } y = -1, \text{ and } y \neq -x, y \neq 1.
\end{array}
$$

Thus, $\frac{x y + 1}{x + y} = -1$, and $y \neq 1$
$\Rightarrow x y + x + y + 1 = 0$, and $y \neq -x, y \neq 1$
$\Rightarrow x = -1$ or $y = -1$, and $y \neq -x, y \neq 1$.
Therefore, when $x = -1$ or $y = -1$, and $y \neq -x$, take $n_{0} = 2$. When $n \geqslant 2$, $a_{n}$ is always the constant -1. -1, but when $n \geqslant 0$, $a_{n}$ is always a constant, so
$a_{n} \neq -1 (n \geqslant n_{0}, n_{0} \geqslant 2)$.
Then $\frac{a_{n} - 1}{a_{n} + 1} = \left(\frac{y - 1}{y + 1}\right)^{F_{n-1}} \left(\frac{x - 1}{x + 1}\right)^{F_{n-2}}$ is always a constant when $n \geqslant n_{0}$.
$$
\begin{array}{l}
\text{Clearly, } \frac{x - 1}{x + 1} \neq 1, \frac{y - 1}{y + 1} \neq 1. \\
\text{If } \frac{x - 1}{x + 1} = -1 \text{ and } \frac{y - 1}{y + 1} = -1, \text{ then } x = y = 0,
\end{array}
$$
which makes the denominator of
$$
a_{2} = \frac{a_{0} a_{1} + 1}{a_{0} + a_{1}}
$$
zero, leading to a contradiction. Therefore, it can only be $\frac{x - 1}{x + 1} = 0$ or $\frac{y - 1}{y + 1} = 0$, i.e., $x = 1$ or $y = 1$, and $y \neq -x$, when $n \geqslant n_{0} (n_{0} \geqslant 2)$, $a_{n}$ is always the constant 1.

In summary, when $|x| = 1$ and $y \neq -x$ or $|y| = 1$ and $x \neq -y$, there exists a positive integer $n_{0}$ such that when $n \geqslant n_{0}$, $a_{n}$ is always the constant 1 or -1.
(2) Notice that
$\frac{a_{n} - 1}{a_{n} + 1} = \left(\frac{y - 1}{y + 1}\right)^{F_{n-1}} \left(\frac{x - 1}{x + 1}\right)^{F_{n-2}} (n \geqslant 2)$.
Then $a_{n} = \frac{2}{1 - \left(\frac{y - 1}{y + 1}\right)^{F_{n-1}} \left(\frac{x - 1}{x + 1}\right)^{F_{n-2}}} - 1$
$= \frac{2(y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}}}{(y + 1)^{F_{n-1}} (x + 1)^{F_{n-2}} - (y - 1)^{F_{n-1}} (x - 1)^{F_{n-2}}} - 1$. $a_{0} = x, a_{1} = y$.