Example 11 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 $a_{n}$.
(2006 National High School League Question)
(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 each 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 this $n=1$, we get $|y|=1$ and $x \neq -y$.
If this $n>1$, we have $a_{n}-1=\frac{a_{n-1} a_{n-2}+1}{a_{n-1}+a_{n-2}}-1=\frac{(a_{n-1}-1)(a_{n-2}-1)}{a_{n-1}+a_{n-2}}, n \geqslant 2$ and $a_{n}+1=\frac{a_{n-1} a_{n-2}+1}{a_{n-1}+a_{n-2}}+1=\frac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-1}+a_{n-2}}, n \geqslant 2$.
Multiplying both sides of the two equations, 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 the above formula, it must be that (1) or $|x|=1$ and $y \neq -x$.
Conversely, if (1) or the above formula is satisfied, then when $n \geqslant 2$, it must be that $a_{n}=$ constant and the constant is 1 or -1.
(2) From the two equations in (2), we can 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 when $n \geqslant 2$, $b_{n}=b_{n-1} b_{n-2}=(b_{n-2} \cdot b_{n-3}) \cdot b_{n-2}=b_{n-2}^{2} \cdot b_{n-3}=$ $\left(b_{n-3} b_{n-4}\right)^{2} b_{n-3}=b_{n-3}^{3} b_{n-4}^{2}=\cdots$.

By this recursion, we get $\frac{a_{n}-1}{a_{n}+1}=\left(\frac{y-1}{y+1}\right)_{n-1}^{F_{n}} \cdot\left(\frac{x-1}{x+1}\right)^{F_{n-1}}, n \geqslant 2$. Where $F_{n}=F_{n-1}+F_{n-2}, n \geqslant 2, F_{0}=F_{1}=1$.
From the above, we find $F_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$, and $n$ can also be extended to have $F_{-1}=0, F_{-2}=1$.
Thus, (3) holds for all $n \geqslant 0$, and from (3) we solve
$$
a_{n}=\frac{(x+1)^{F_{n-2}} \cdot(y+1)^{F_{n-1}}+(x+1)^{F_{n-2}} \cdot(y-1)^{F_{n-1}}}{(x+1)^{F_{n-2}} \cdot(y+1)^{F_{n-1}}-(x-1)^{F_{n-2}} \cdot(y-1)^{F_{n-1}}}, n \geqslant 0 .
$$