Example 4 Given that $x, y, z$ are real numbers, and satisfy
$$x+y+z=xy+yz+zx \text{.}$$

Find the minimum value of $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}$.
Let $x=1, y=z=-1$. Then
$$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}=-\frac{1}{2}$$

Conjecture the minimum value is $-\frac{1}{2}$.
It suffices to prove:
$$\begin{aligned}
& \frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1} \geqslant-\frac{1}{2} \\
\Leftrightarrow & \frac{(x+1)^{2}}{x^{2}+1}+\frac{(y+1)^{2}}{y^{2}+1} \geqslant \frac{(z-1)^{2}}{z^{2}+1}
\end{aligned}$$

Notice that $z(x+y-1)=x+y-xy$.
If $x+y-1=0$, then $x+y=xy=1$. Contradiction.
Hence, $x+y-1 \neq 0$.
Thus, $z=\frac{x+y-xy}{x+y-1}$.
Substituting into inequality (39) gives
$$\begin{aligned}
& \frac{(x+1)^{2}}{x^{2}+1}+\frac{(y+1)^{2}}{y^{2}+1} \\
\geqslant & \frac{(xy-1)^{2}}{(x+y-1)^{2}+(x+y-xy)^{2}}
\end{aligned}$$

By Cauchy-Schwarz inequality, the left side of (40)
$$\begin{array}{l}
\geqslant \frac{[(1+x)(1-y)+(1+y)(1-x)]^{2}}{\left(1+x^{2}\right)(1-y)^{2}+\left(1+y^{2}\right)(1-x)^{2}} \\
=\frac{4(xy-1)^{2}}{\left(1+x^{2}\right)(1-y)^{2}+\left(1+y^{2}\right)(1-x)^{2}}
\end{array}$$

Thus, it suffices to prove
$$\begin{aligned}
& 4(x+y-1)^{2}+4(x+y-xy)^{2} \\
\geqslant & \left(1+x^{2}\right)(1-y)^{2}+\left(1+y^{2}\right)(1-x)^{2} \\
\Leftrightarrow & f(x)=\left(y^{2}-3 y+3\right) x^{2}-\left(3 y^{2}-8 y+3\right) x+3 y^{2}-3 y+1 \geqslant 0
\end{aligned}$$

By $\Delta=\left(3 y^{2}-8 y+3\right)^{2}-4\left(y^{2}-3 y+3\right)\left(3 y^{2}-3 y+1\right)=$ $-3\left(y^{2}-1\right)^{2} \leqslant 0$, hence $f(x) \geqslant 0$ always holds.

Therefore, the conjecture is true, i.e., the minimum value of $\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}$ is $-\frac{1}{2}$.