Example 8 The maximum value of the function $f(x)=\sqrt{x^{4}-3 x^{2}-6 x+13}$ $-\sqrt{x^{4}-x^{2}+1}$ is $\qquad$ .
(1992, National High School Mathematics Competition)
Given $f(x)=\sqrt{x^{4}-3 x^{2}-6 x+13}$
$$
\begin{aligned}
& -\sqrt{x^{4}-x^{2}+1} \\
= & \sqrt{(x-3)^{2}-\left(x^{2}-2\right)^{2}}-\sqrt{x^{2}+\left(x^{2}-1\right)^{2}},
\end{aligned}
$$

we can see that the geometric meaning of the function $y=f(x)$ is: as shown in Figure 5, for a point $P\left(x, x^{2}\right)$ on the parabola $y=x^{2}$, the difference in distances from point $P$ to point $A$ $(3,2)$ and point $B(0,1)$.
Since point $A$ is below the parabola and point $B$ is above the parabola, the line $AB$ intersects the parabola. The intersection points are determined by the system of equations $\left\{\begin{array}{l}y=x^{2}, \\ \frac{y-1}{x-0}=\frac{2-1}{3-0}\end{array}\right.$. Eliminating $y$ yields $3 x^{2}-x-3=0$. Since the constant term of this equation is negative, the equation must have a negative root.
Because the difference of two sides of a triangle is less than the third side, when point $P$ is at the intersection point $C$ corresponding to the negative root, $f(x)$ has a maximum value of $|AB|=\sqrt{10}$.