Example 5 In the sequence $\left\{x_{n}\right\}$, $x_{1}=3, x_{2}=2, x_{n}=\frac{x_{n-2} x_{n-1}}{2 x_{n-2}-x_{n-1}}(n \geqslant 3, n \in \mathbf{N})$, find the general term formula of the sequence $\left\{x_{n}\right\}$.
Solution: Taking the reciprocal of the original recursive formula, we get $\frac{1}{x_{n}}=\frac{2}{x_{n-1}}-\frac{1}{x_{n-2}}(n \geqslant 3)$.
That is, $\frac{1}{x_{n}}-\frac{1}{x_{n-1}}=\frac{1}{x_{n-1}}-\frac{1}{x_{n-2}}=\cdots=\frac{1}{x_{2}}-\frac{1}{x_{1}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.
Therefore, $\frac{1}{x_{n}}-\frac{1}{x_{n-1}}=\frac{1}{6}(n \geqslant 2)$.
Thus, $\frac{1}{x_{n}}=\left(\frac{1}{x_{n}}-\frac{1}{x_{n-1}}\right)+\left(\frac{1}{x_{n-1}}-\frac{1}{x_{n-2}}\right)+\cdots+\left(\frac{1}{x_{2}}-\frac{1}{x_{1}}\right)+\frac{1}{x_{1}}=\frac{1}{x_{1}}+\frac{1}{6} \cdot(n-1)$ $=\frac{1}{3}+\frac{n-1}{6}=\frac{n+1}{6}$.
Hence, $x_{n}=\frac{6}{n+1}\left(n \in \mathbf{N}^{*}\right)$.
Analysis: This problem employs the method of successive differences and the reciprocal method, especially the use of the reciprocal method, which simplifies the seemingly intractable formula. When solving problems like $a_{n+1}=\frac{a_{n}}{\alpha a_{n}+\beta}$ or $a_{n+1}=\frac{a_{n} a_{n-1}}{\alpha a_{n}+\beta a_{n-1}}$, the reciprocal method often plays a role in simplifying the problem.