2. Find all real solutions of the system of equations
$$
\left\{\begin{array}{l}
5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right), \\
x y+y z+z x=1
\end{array}\right.
$$

(Supplied by Zhu Huawei)
2. From $x+\frac{1}{x}=\frac{x^{2}+x y+y z+z x}{x}$
$$
=\frac{(x+y)(x+z)}{x},
$$

the first equation can be transformed into
$$
\begin{array}{l}
5 y z(x+y)(x+z)=12 x z(y+x)(y+z) \\
=13 x y(z+x)(z+y) . \\
\text { Let }\left\{\begin{array}{l}
x(y+z)=a, \\
y(z+x)=b, \\
z(x+y)=c .
\end{array}\right.
\end{array}
$$

then $a+b+c=2$,
$5 b c=12 c a=13 a b$
$$
\Leftrightarrow \frac{5}{a}=\frac{12}{b}=\frac{13}{c}=k \text {. }
$$

Substituting (2) into (1) yields
$$
\frac{5}{k}+\frac{12}{k}+\frac{13}{k}=2 \text {. }
$$

Solving for $k$ gives $k=15$.
Then, from $x y+x z=\frac{5}{15}, y z+x y=\frac{12}{15}, z x+z y=\frac{13}{15}$, we get
$$
y z=\frac{10}{15}, z x=\frac{3}{15}, x y=\frac{2}{15} .
$$

Therefore, $x y z= \pm \sqrt{\frac{10}{15} \times \frac{3}{15} \times \frac{2}{15}}= \pm \frac{2}{15}$.
Thus, the original system of equations has two solutions:
$$
\left(\frac{1}{5}, \frac{2}{3}, 1\right) \text { and }\left(-\frac{1}{5},-\frac{2}{3},-1\right) \text {. }
$$