Example 1 Solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{1}{x y}=\frac{x}{z}+1, \\
\frac{1}{y z}=\frac{y}{x}+1, \\
\frac{1}{z x}=\frac{z}{y}+1 .
\end{array}\right.
$$
Solving the first two equations by subtraction, we get
$$
\begin{array}{l}
\frac{z-x}{x y z}=\frac{x}{z}-\frac{y}{x}=\frac{x^{2} y-y^{2} z}{x y z} \\
\Rightarrow z-x=x^{2} y-y^{2} z \\
\Rightarrow z\left(1+y^{2}\right)=x(x y+1) .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
x\left(1+z^{2}\right)=y(y z+1), \\
y\left(1+x^{2}\right)=z(z x+1) .
\end{array}
$$

Then, \( x y z\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right) \)
$$
=x y z(x y+1)(y z+1)(z x+1) \text {. }
$$

Since \( x y z \neq 0 \), we have
$$
\begin{array}{l}
\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right) \\
=(x y+1)(y z+1)(z x+1) .
\end{array}
$$

Expanding and simplifying, we get
$$
\begin{array}{l} 
x^{2}+y^{2}+z^{2}+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \\
= x y+y z+z x+x^{2} y z+y^{2} z x+z^{2} x y \\
\Rightarrow \frac{1}{2}\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}+\right. \\
\left.(x y-y z)^{2}+(y z-z x)^{2}+(z x-x y)^{2}\right) \\
\quad=0 \\
\Rightarrow x=y=z .
\end{array}
$$

Therefore, the real solutions to the original system of equations are
$$
(x, y, z)=\left( \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2}\right)
$$