Example 8 Given the infinite sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=x, a_{1}=y, a_{n+1}=\frac{a_{n} a_{n-1}+1}{a_{n}+a_{n-1}}, n=1,2, \cdots$,
(1) For which real numbers $x$ and $y$, 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 $a_{n}$.
Solution: (1) We have
$$
a_{n}-a_{n+1}=a_{n}-\frac{a_{n} a_{n-1}+1}{a_{n}+a_{n-1}}=\frac{a_{n}^{2}-1}{a_{n}+a_{n-1}}, n=1,2, \cdots .
$$

Therefore, if for some positive integer $n$, we have $a_{n+1}=a_{n}$, then it must be that $a_{n}^{2}=1$, and $a_{n}+a_{n-1} \neq 0$.
If $n=1$, we get
$$
|y|=1 \quad \text { and } x \neq -y \text {. }
$$

If $n>1$, we have
$$
a_{n}-1=\frac{a_{n-1} a_{n-2}+1}{a_{n-1}+a_{n-2}}-1=\frac{\left(a_{n-1}-1\right)\left(a_{n-2}-1\right)}{a_{n-1}+a_{n-2}}, n \geqslant 2,
$$

and
$$
a_{n}+1=\frac{a_{n-1} a_{n-2}+1}{a_{n-1}+a_{n-2}}+1=\frac{\left(a_{n-1}+1\right)\left(a_{n-2}+1\right)}{a_{n-1}+a_{n-2}}, n \geqslant 2,
$$

Multiplying both sides of (3) and (4), we get
$$
a_{n}^{2}-1=\frac{a_{n-1}^{2}-1}{a_{n-1}+a_{n-2}} \cdot \frac{a_{n-2}^{2}-1}{a_{n-1}+a_{n-2}}, n \geqslant 2,
$$

By (5) recursion, we must have (2) or
$$
|x|=1 \quad \text { and } y \neq \cdots x \text {. }
$$

Conversely, if condition (2) or (3) is satisfied, then for $n \geqslant 2$, we must have $a_{n}=$ constant, and the constant is 1 or -1.
(2) From (3) and (4), we get
$$
\frac{a_{n}-1}{a_{n}+1}=\frac{a_{n-1}-1}{a_{n-1}+1} \cdot \frac{a_{n-2}-1}{a_{n-2}+1}, n \geqslant 2 .
$$

Let $b_{n}=\frac{a_{n}-1}{a_{n}+1}$, then for $n \geqslant 2$,
$$
b_{n}=b_{n-1} b_{n-2}=\left(b_{n-2} b_{n-3}\right) b_{n-2}=b_{n-2}^{2} b_{n-3}=\left(b_{n-3} b_{n-4}\right)^{2} b_{n-3}=b_{n \cdot 3}^{3} b_{n-4}^{2}=\cdots
$$

By this recursion, we get
$$
\frac{a_{n}-1}{a_{n}+1}=\left(\frac{y-1}{y+1}\right)^{F_{n-1}} \cdot\left(\frac{x-1}{x+1}\right)^{F_{n-2}}, n \geqslant 2,
$$

where
$$
F_{n}=F_{n-1}+F_{n-2}, n \geqslant 2, F_{0}=F_{1}=1 .
$$

From (2.9), we get
$$
F_{n}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\right) .
$$

The $n$ in the above formula can also be extended to negative values, for example
$$
F_{-1}=0, F_{-2}=1 .
$$

In this way, (8) holds for all $n \geqslant 0$. From (8), we get
$$
a_{n}=\frac{(x+1)^{F_{n-2}}(y+1)^{F_{n-1}}+(x-1)^{F_{n-2}}(y-1)^{F_{n-1}}}{(x+1)^{F_{n-2}}(y+1)^{F_{n}}-(x-1)^{F_{n-2}}(y-1)^{F_{n-1}}}, n \geqslant 0 .
$$
In (11), $F_{n-1}, F_{n-2}$ are determined by (10).