Example 1 Given the sequence $\left\{a_{n}\right\}$,
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, a_{3}=-1, \\
a_{n+2} a_{n+1}=4 a_{n+1} a_{n}-6 a_{n+1} a_{n-1}+9 a_{n} a_{n-1}-6 a_{n}^{2} .
\end{array}
$$

Find $a_{n}$.
Analysis: This recursive sequence is a nonlinear homogeneous recursive sequence and cannot be directly solved using the characteristic root method. Note that the recursive formula is a quadratic homogeneous form, which can be transformed into a linear homogeneous form through factorization.
Solution: Since
$$
\begin{array}{l}
a_{n+2} a_{n+1}=4 a_{n+1} a_{n}-6 a_{n+1} a_{n-1}+9 a_{n} a_{n-1}-6 a_{n}^{2} \\
=\left(2 a_{n+1}-3 a_{n}\right)\left(2 a_{n}-3 a_{n-1}\right),
\end{array}
$$

then $\frac{a_{n+2}}{2 a_{n+1}-3 a_{n}}=\left(\frac{a_{n+1}}{2 a_{n}-3 a_{n-1}}\right)^{-1}=\cdots$
$$
=\left(\frac{a_{3}}{2 a_{2}-3 a_{1}}\right)^{(-1)^{n-1}}=-1 \text {. }
$$

Thus, $a_{n+2}=-2 a_{n+1}+3 a_{n}$.
The characteristic equation is $x^{2}=-2 x+3$.
Solving it, we get $x_{1}=1, x_{2}=-3$.
Assume $a_{n}=A+B(-3)^{n}$, substituting the initial values we have

Therefore, $a_{n}=\frac{5}{4}+\frac{1}{12}(-3)^{n}$.