$$
\left\{\begin{array}{l}
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right), \\
x y+y z+z x=1
\end{array}\right.
$$

Find all real solutions to the system of equations.
Transform equation (1) into
$$
\frac{x}{5\left(1+x^{2}\right)}=\frac{y}{12\left(1+y^{2}\right)}=\frac{z}{13\left(1+z^{2}\right)} \text {. }
$$

From equation (3), it is known that $x$, $y$, and $z$ have the same sign.
From equation (2), we get $x=\frac{1-y z}{y+z}$.
Substituting the above equation into equation (1) yields
$$
\begin{array}{l}
12\left(\frac{y^{2}+1}{y}\right)=5\left(\frac{1-y z}{y+z}+\frac{y+z}{1-y z}\right) \\
=5 \frac{(1-y z)^{2}+(y+z)^{2}}{(y+z)(1-y z)}=5 \frac{\left(y^{2}+1\right)\left(z^{2}+1\right)}{(y+z)(1-y z)} .
\end{array}
$$

Thus, $5\left(z^{2}+1\right) y=12(y+z)(1-y z)$.
Similarly, $5\left(y^{2}+1\right) z=13(y+z)(1-y z)$.
Adding the two equations gives
$$
\left(z^{2}+1\right) y+\left(y^{2}+1\right) z=5(y+z)(1-y z) \text {. }
$$

Solving this, we get $y=\frac{2}{3 z}$.
Substituting $y=\frac{2}{3 z}$ into the last equation of (1) yields
$$
z= \pm 1, y= \pm \frac{2}{3} \text {. }
$$

Consequently, $x= \pm \frac{1}{5}$.
Therefore, the system of equations has two solutions
$$
\left(\frac{1}{5}, \frac{2}{3}, 1\right),\left(-\frac{1}{5},-\frac{2}{3},-1\right) \text {. }
$$