Example 3 Let the three sides of $\triangle A B C$ be $a, b$, and $c$, and $\frac{a-b}{1+a b}+\frac{b-c}{1+b c}+\frac{c-a}{1+c a}=0$. Then the shape of $\triangle A B C$ must be a $\qquad$ triangle.
Solution: Remove the denominator from the original expression and set it as $f$, we get
$$
\begin{aligned}
f= & (a-b)(1+b c)(1+c a)+(b-c)(1+a b) . \\
& (1+c a)+(c-a)(1+b c)(1+a b) \\
= & a\left(b^{2}-c^{2}\right)+b\left(c^{2}-a^{2}\right)+c\left(a^{2}-b^{2}\right) \\
= & 0 .
\end{aligned}
$$

When $a=b$, $f=0$, by the factor theorem, $f$ has a factor $a-b$. Since $f$ is a cyclic symmetric expression in $a, b, c$, by its properties, $f$ also has factors $b-c$ and $c-a$. Therefore, $f$ has a factor
$$
g=(a-b)(b-c)(c-a) .
$$

Since both $f$ and $g$ are third-degree homogeneous cyclic symmetric expressions, $f$ and $g$ differ only by a non-zero constant factor, i.e.,
$$
f=k(a-b)(b-c)(c-a)=0 .
$$

This implies that at least one of $a-b, b-c, c-a$ is equal to 0, i.e., at least two of $a, b, c$ are equal, so the triangle has at least two equal sides.
Therefore, the triangle is isosceles.