Example 11 (16th "Hope Cup" Senior High School Competition Problem) In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n}-a_{n-1}=a_{n+1}+3 a_{n} a_{n+1}$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Given that $y=f(x)$ is an even function, and for any $x$ there is $f(1+x)=f(1-x)$, when $x \in[-2,-1)$, $f(x)=\log _{2}|x+1|$. Find the range of $x$ that makes $f(x)+a_{n}<0\left(n \in \mathbf{N}^{*}\right)$ always true.
Solve (1) Since
Therefore
$$
\begin{array}{c}
a_{n}-a_{n-1}=a_{n-1}+3 a_{n} a_{n-1}, \\
a_{n}=2 a_{n-1}+3 a_{n} a_{n-1}=a_{n \cdot 1}\left(2+3 a_{n}\right), \\
\frac{1}{a_{n-1}}=\frac{2}{a_{n}}+3,
\end{array}
$$

Thus
So $\frac{1}{a_{n+1}}+3=2\left(\frac{1}{a_{n}}+3\right), a_{1}=1,\left\{\frac{1}{a_{n}}+3\right\}$ is a geometric sequence with the first term 4 and common ratio 2, so
$$
\frac{1}{a_{n}}+3=4 \times 2^{n} \quad 1=2^{n+1},
$$

Thus
$$
a_{n}=\frac{1}{2^{n+1}-3}\left(n \in \mathbf{N}^{-}\right) .
$$
(2) Since $f(x+2)=f[1+(x+1)]=f[1-(x+1)]=f(-x)=f(x)$. Therefore, $f(x)$ is a periodic function with a period of 2. First, find the solution set of the inequality $f(x)+a_{n}<0$ when $x \in(-1,1)$.
For $x \in[0,1) \| 1, x-2 \in[-2,-1)$.
So $f(x)=f(x-2)=\log _{2}|(x-2)+1|=\log _{2}|x-1|=\log _{2}(1-x)$.
Since $a_{0}=\frac{1}{2^{n}-3}$.
Therefore, $\left\{a_{n}\right\}$ is a monotonically decreasing sequence.
Hence $a_{n} \leqslant a_{1}=1,-a_{n} \geqslant-1$.
To make $f(x)+a_{n}<0$, i.e., $f(x)<-a_{n}$ hold for all $x \in [0,1)$, it is sufficient that $f(x)<-1$, i.e.,
$$
\log _{2}(1-x)<-1 \text {, }
$$

Solving this gives
$$
\frac{1}{2}<x<1 \text {. }
$$

Since $y=f(x)$ is an even function, for $x \in(-1,0)$, $f(x)=\log _{2}(1+x)$,
Similarly, $\log _{2}(1+x)<-1$,
Solving this gives $-1<x<-\frac{1}{2}$.
Since $f(x)$ is a periodic function with a period of 2, the range of $x$ that makes $f(x)+a_{n}<0\left(n \in \mathbf{N}^{*}\right)$ hold is
$$
\left\{x \left\lvert\,-1+2 k<x<-\frac{1}{2}+2 k\right., k \in \mathbf{Z}\right\} \text { or }\left\{x \left\lvert\, \frac{1}{2}+2 k<x<1+2 k\right., k \in \mathbf{Z}\right\} \text {. }
$$