Example 12 (Problem from the 4th China Girls Mathematical Olympiad) Solve the system of equations
$$
\left\{\begin{array}{l}
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right) \\
x y+y z+z x=1
\end{array}\right.
$$
Solution 1 (1) can be transformed into
$$
\frac{x}{5\left(1+x^{2}\right)}=\frac{y}{12\left(1+y^{2}\right)}=\frac{z}{13\left(1+z^{2}\right)} .
$$
(3)
It is clear that $x, y, z$ have the same sign. First, we find the positive solutions.
There exist $\alpha, \beta, \gamma \in(0, \pi)$, such that $x=\tan \frac{\alpha}{2}, y=\tan \frac{\beta}{2}, z=\tan \frac{\gamma}{2}$,
then $\sin \alpha=\frac{2 x}{1+x^{2}}, \sin \beta=\frac{2 y}{1+y^{2}}, \sin \gamma=\frac{2 z}{1+z^{2}}$,
(3) is $\frac{\sin \alpha}{5}=\frac{\sin \beta}{12}=\frac{\sin \gamma}{13}$.
(2) can be transformed into $\frac{1}{z}=\frac{x+y}{1-x y}$, i.e., $\cot \frac{\gamma}{2}=\tan \frac{\alpha+\beta}{2}$.

Note that $z \neq 0, x y \neq 1$, because $\alpha, \beta, \gamma \in(0, \pi)$, so $\frac{\alpha+\beta}{2}=\frac{\pi}{2}-\frac{\gamma}{2}$,
i.e., $\alpha+\beta+\gamma=\pi$.
Thus, $\alpha, \beta, \gamma$ are the three interior angles of some triangle $A B C$.
By (4) and the Law of Sines, the ratio of the sides $a, b, c$ opposite to $\alpha, \beta, \gamma$ is $5: 12: 13$,
so, $\sin \alpha=\frac{5}{13}, \sin \beta=\frac{12}{13}, \sin \gamma=1$.
Thus, $x=\tan \frac{\alpha}{2}=\frac{1}{5}$ or $5, y=\tan \frac{\beta}{2}=\frac{2}{3}$ or $\frac{3}{2}, z=\tan \frac{\gamma}{2}=1$.
Substituting $z=1$ into (2), it is easy to see that $x$ and $y$ are both less than 1. Therefore, $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ is the only positive solution.
Thus, the original system of equations has two solutions: $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ and $\left(-\frac{1}{5},-\frac{2}{3},-1\right)$.
Solution 2 It is clear that $x, y, z$ have the same sign.
From (2), we get $x=\frac{1-y z}{y+z}$, substituting into (1) we get
$$
\begin{aligned}
12\left(\frac{y^{2}+1}{y}\right) & =5\left(\frac{1-y z}{y+z}+\frac{y+z}{1-y z}\right) \\
& =5 \frac{(1-y z)^{2}+(y+z)^{2}}{(y+z)(1-y z)}=5 \frac{\left(y^{2}+1\right)\left(z^{2}+1\right)}{(y+z)(1-y z)},
\end{aligned}
$$

i.e., $5\left(z^{2}+1\right) y=12(y+z)(1-y z)$.
Similarly, $5\left(y^{2}+1\right) z=13(y+z)(1-y z)$,
rearranging gives $12 y^{2} z+17 y z^{2}=7 y+12 z, 18 y^{2} z+13 y z^{2}=13 y+8 z$,
adding the two equations, we get $30 y z(y+z)=20(y+z)$,
so $y z=\frac{2}{3}, y=\frac{2}{3 z}$, substituting into (1) we solve to get $z= \pm 1$.
Thus, the original system of equations has two solutions: $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ and $\left(-\frac{1}{5},-\frac{2}{3},-1\right)$.
Solution 3 $x+\frac{1}{x}=\frac{x^{2}+x y+y z+z x}{x}=\frac{(x+y)(x+z)}{x}$,
so (1) can be transformed into $5 y z(x+y)(x+z)=12 x z(y+x)(y+z)$
$$
=13 x y(z+x)(z+y)
$$

Let $\left\{\begin{array}{l}x(y+z)=a, \\ y(z+x)=b, \\ z(x+y)=c,\end{array}\right.$
then from (2) we get $a+b+c=2$,
(5) becomes $5 b c=12 c a=13 a b \Leftrightarrow \frac{5}{a}=\frac{12}{b}=\frac{13}{c}=k$.
(7) substituting into (6), we get $\frac{5}{k}+\frac{12}{k}+\frac{13}{k}=2$, so $k=15$.

Thus, $\left\{\begin{array}{l}x y+x z=\frac{5}{15}, \\ y z+x y=\frac{12}{15} \\ z x+z y=\frac{13}{15}\end{array} \Rightarrow\left\{\begin{array}{l}y z=\frac{10}{15}, \\ z x=\frac{3}{15}, \\ x y=\frac{2}{15} .\end{array}\right.\right.$
Multiplying the three equations, we get $\quad x y z= \pm \sqrt{\frac{10}{15} \cdot \frac{3}{15} \cdot \frac{2}{15}}= \pm \frac{2}{15}$,
Thus, the original system of equations has two solutions: $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ and $\left(-\frac{1}{5},-\frac{2}{3},-1\right)$.
Solution 4 Let $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)=k$, then
$$
\frac{k}{x+\frac{1}{x}}=5, \frac{k}{y+\frac{1}{y}}=12, \frac{k}{z+\frac{1}{z}}=13 .
$$

So $\frac{1}{\left(x+\frac{1}{x}\right)^{2}}+\frac{1}{\left(y+\frac{1}{y}\right)^{2}}=\frac{1}{\left(z+\frac{1}{z}\right)^{2}}$,
i.e., $\left(\frac{x}{x^{2}+1}\right)^{2}+\left(\frac{y}{y^{2}+1}\right)^{2}=\left(\frac{z}{z^{2}+1}\right)^{2}$.
Also, $x^{2}+1=x^{2}+x y+y z+z x=(x+y)(x+z)$,
so (8) can be transformed into
$$
\frac{x^{2}}{(x+y)^{2}(x+z)^{2}}+\frac{y^{2}}{(y+z)^{2}(y+x)^{2}}=\frac{z^{2}}{(x+z)^{2}(y+z)^{2}},
$$

i.e., $2 x^{2} y^{2}+2 x^{2} y z+2 x y^{2} z=2 x y z^{2}$,
$$
2 x y(x y+y z+z x)=2 x y z^{2} \text {, }
$$
$$
x y=x y z^{2} \text {. }
$$

Since $x, y, z \neq 0$, so $z^{2}=1$. The two solutions of the original equation are $\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ and $\left(-\frac{1}{5},-\frac{2}{3},-1\right)$.