Example 6 Solve the system of equations in the real numbers
$$
\left\{\begin{array}{l}
x y+y z+z x=1, \\
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right) .
\end{array}\right.
$$
From equation (2), we know that $x, y, z$ have the same sign.
Therefore, we can first find the positive solutions.
In $\triangle ABC$, let the sides opposite to $\angle A, \angle B, \angle C$ be $a, b, c$ respectively.
From equation (1), we recall that in $\triangle ABC$, there is the identity
$$
\tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} + \tan \frac{A}{2} \cdot \tan \frac{B}{2} = 1.
$$

Thus, we can set $x = \tan \frac{A}{2}, y = \tan \frac{B}{2}, z = \tan \frac{C}{2}$.
Substituting into equation (2) and simplifying, we get
$$
\begin{array}{l}
\sin A : \sin B : \sin C = 5 : 12 : 13 \\
\Rightarrow \cos A = \frac{12}{13}, \cos B = \frac{5}{13}, \cos C = 0 \\
\Rightarrow x = \tan \frac{A}{2} = \frac{1}{5}, y = \tan \frac{B}{2} = \frac{2}{3}, z = \tan \frac{C}{2} = 1. \\
\text{Therefore, } (x, y, z) = \left(\frac{1}{5}, \frac{2}{3}, 1\right) \text{ and } \left(-\frac{1}{5}, -\frac{2}{3}, -1\right).
\end{array}
$$