9. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=\frac{3 a_{n}-4}{9 a_{n}+15}\left(n \in N^{*}\right)$, find the general term $a_{n}$ of the sequence $\left\{a_{n}\right\}$.
Solution: Consider the fixed point $x, \because a_{n+1}=\frac{3 a_{n}-4}{9 a_{n}+15}, \therefore x=\frac{3 x-4}{9 x+15}$,
i.e., $9 x^{2}+12 x+4=0, \therefore x_{1,2}=-\frac{2}{3}, \therefore a_{n+1}+\frac{2}{3}=\frac{3 a_{n}-4}{9 a_{n}+15}+\frac{2}{3}=\frac{3 a_{n}+2}{3 a_{n}+5}$,
i.e., $\frac{1}{a_{n+1}+\frac{2}{3}}=\frac{3\left(a_{n}+\frac{2}{3}\right)+3}{3\left(a_{n}+\frac{2}{3}\right)}=\frac{1}{a_{n}+\frac{2}{3}}+1$, let $b_{n}=\frac{1}{a_{n}+\frac{2}{3}}$, and $b_{1}=\frac{1}{3+\frac{2}{3}}=\frac{3}{11}$,
$\therefore b_{n+1}=b_{n}+1, \quad \therefore b_{n}=\frac{3}{11}+n-1=n-\frac{8}{11}$,
$\mid \therefore$ The general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=\frac{49-22 n}{3(11 n-8)}$