Example 4 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{n+1}=\frac{3 a_{n}-1}{4 a_{n}+7} \text {. }
$$

Find the general term formula of the sequence $\left\{a_{n}\right\}$.
Let $x=\frac{3 x-1}{4 x+7}$. Then
$$
4 x^{2}+4 x+1=0 \text {. }
$$

Therefore, $x_{0}=-\frac{1}{2}$ is a fixed point of the function $f(x)=\frac{3 x-1}{4 x+7}$.
Hence, $a_{n+1}+\frac{1}{2}=\frac{5\left(a_{n}+\frac{1}{2}\right)}{4 a_{n}+7}$.
Then $\frac{1}{a_{n+1}+\frac{1}{2}}=\frac{\frac{4}{5}\left(a_{n}+\frac{1}{2}\right)+1}{a_{n}+\frac{1}{2}}$
$$
=\frac{1}{a_{n}+\frac{1}{2}}+\frac{4}{5} \text {. }
$$

Therefore, the sequence $\left\{\frac{1}{a_{n}+\frac{1}{2}}\right\}$ is an arithmetic sequence with the first term $\frac{1}{a_{1}+\frac{1}{2}}=1$ and common difference $\frac{4}{5}$.
Thus, $\frac{1}{a_{n}+\frac{1}{2}}=\frac{1}{a_{1}+\frac{1}{2}}+(n-1) \frac{4}{5}=\frac{4}{5} n+\frac{1}{5}$.
Therefore, $a_{n}=\frac{9-4 n}{2+8 n}$.