9. (16 points) Given the positive sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
\sqrt{a_{n} a_{n+1}+a_{n} a_{n+2}} \\
=4 \sqrt{a_{n} a_{n+1}+a_{n+1}^{2}}+3 \sqrt{a_{n} a_{n+1}}
\end{array}
$$

and $a_{1}=1, a_{2}=8$.
Find the general term formula of $\left\{a_{n}\right\}$.
$$
\begin{array}{l}
\sqrt{1+\frac{a_{n+2}}{a_{n+1}}}=4 \sqrt{1+\frac{a_{n+1}}{a_{n}}}+3 \\
\Rightarrow \sqrt{1+\frac{a_{n+2}}{a_{n+1}}}+1=4\left(\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1\right) . \\
\text { Let } b_{n}=\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1 .
\end{array}
$$

Then $b_{1}=4, b_{n+1}=4 b_{n}$.
Thus, $b_{n}=4^{n-1} b_{1}=4^{n}$.
Therefore, $a_{n+1}=\left[\left(4^{n}-1\right)^{2}-1\right] a_{n}$.
Hence, when $n>1$,
$$
\begin{array}{l}
a_{n}=\prod_{k=1}^{n-1}\left[\left(4^{k}-1\right)^{2}-1\right] a_{1} \\
=\prod_{k=1}^{n-1}\left[\left(4^{k}-1\right)^{2}-1\right] .
\end{array}
$$

Thus, $a_{n}=\left\{\begin{array}{ll}1, & n=1 ; \\ \prod_{k=1}^{n-1}\left[\left(4^{k}-1\right)^{2}-1\right], & n \geqslant 2 \text {. }\end{array}\right.$