Example 3 If there exist two points on the parabola $y=x^{2}$ that are symmetric about the line $y=m(x-3)$, find the range of values for $m$.
As shown in Figure 8-3, let points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola be symmetric about the line $y=m(x-3)$, with the midpoint of $A B$ being $M\left(x_{0}, y_{0}\right)$. Clearly, $m \neq 0$.
$$
\begin{array}{l}
k_{A H}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{x_{2}^{2}-x_{1}^{2}}{x_{2}-x_{1}}=x_{2}+x_{1}=-\frac{1}{m}=2 x_{0}, \\
x_{0}=-\frac{1}{2 m}, y_{0}=\frac{y_{1}+y_{2}}{2}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{2}=\frac{\frac{1}{m^{2}}-2 x_{1} x_{2}}{2} .
\end{array}
$$

Since $M\left(x_{0}, y_{0}\right)$ lies on the line $y=m(\dot{x}-3)$, we have
$$
x_{1} x_{2}=\frac{1}{2 m^{2}}+\frac{1}{2}+3 m \text {. }
$$

From (1) and (2), we know that $x_{1}, x_{2}$ are the roots of the equation $t^{2}+\frac{1}{m} t+\left(\frac{1}{2 m^{2}}+\frac{1}{2}+3 m\right)=0$. $\Delta>0,\left(x_{1} \neq x_{2}\right),(2 m+1)\left(6 m^{2}-2 m+1\right)<0$, solving this yields $m<-\frac{1}{2}$.