Example 12 (2008 National High School Joint Competition Hubei Province Preliminary Test Question) Let the sequence $\{f(n)\}$ satisfy: $f(1)=1$, $f(2)=2, \frac{f(n+2)}{f(n)}=\frac{f^{2}(n+1)+1}{f^{2}(n)+1} \quad(n \geqslant 1)$.
(1) Find the recurrence relation between $f(n+1)$ and $f(n)$, i.e., $f(n+1)=g[f(n)]$;
(2) Prove: $63<f(2008)<78$.

保留了原文的换行和格式。
Solution (1) Since $f(1)=1, f(2)=2, \frac{f(n+2)}{f(n)}=\frac{f^{2}(n+1)+1}{f^{2}(n)+1}(n \geqslant 1)$, it is easy to see that for all $n \geqslant 1, f(n) \neq 0$.

When $n \geqslant 1$, from $\frac{f(n+2)}{f(n)}=\frac{f^{2}(n+1)+1}{f^{2}(n)+1}$ we can get $\frac{f(n+2) f(n+1)}{f^{2}(n+1)+1}=\frac{f(n+1) f(n)}{f^{2}(n)+1}$, thus $-\frac{f(n+2)}{f(n+1)+\frac{1}{f(n+1)}}=\frac{f(n+1)}{f(n)+\frac{1}{f(n)}}$.

By successively using the above relation, we get $\frac{f(n+1)}{f(n)+\frac{1}{f(n)}}=\frac{f(n)}{f(n-1)+\frac{1}{f(n-1)}}=\frac{f(n-1)}{f(n-2)+\frac{1}{f(n-2)}}=$ $\cdots=\frac{f(2)}{f(1)+\frac{1}{f(1)}}=\frac{2}{1+\frac{1}{1}}=1$, hence $f(n+1)=f(n)+\frac{1}{f(n)}$.
(2) Clearly, from $f(1)=1$ and $f(n+1)=f(n)+\frac{1}{f(n)}(n \geqslant 1)$, we know: for all $n \geqslant 1$, $f(n) \geqslant 1$ holds. Therefore, $06022$, so $63^{2}<f^{2}(2008)<78^{2}$, thus $63<$ $f(2008)<78$.