9. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}$ $=\frac{(n+1) a_{n}^{2}}{2 a_{n}^{2}+4 n a_{n}+n^{2}}$, find the general term of $\left\{a_{n}\right\}$.
Solve:
\[
\frac{a_{n+1}}{n+1}=\frac{\frac{a_{n}^{2}}{n^{2}}}{\frac{2 a_{n}^{2}}{n^{2}}+\frac{4 a_{n}}{n}+1}, \text{ let } b_{n}=\frac{a_{n}}{n}, \text{ then } b_{n+1}=\frac{b_{n}^{2}}{2 b_{n}^{2}+4 b_{n}+1}
\]
\[
\Rightarrow \frac{1}{b_{n+1}}=\frac{1}{b_{n}^{2}}+\frac{4}{b_{n}}+2 \Rightarrow \frac{1}{b_{n+1}}+2=\left(\frac{1}{b_{n}}+2\right)^{2} \Rightarrow \ln \left(\frac{1}{b_{n+1}}+2\right)=2 \ln \left(\frac{1}{b_{n}}+2\right) \text{, }
\]

Let \( c_{n}=\ln \left(\frac{1}{b_{n}}+2\right) \), and \( b_{1}=1, c_{1}=\ln 3 \Rightarrow c_{n}=2^{n-1} \ln 3 \),
so \(\frac{1}{b_{n}}=3^{2^{n-1}}-2 \Rightarrow a_{n}=\frac{n}{3^{2^{n-1}}-2}\left(n \in \mathbf{N}^{*}\right)\).