Example 9 Function
$$
\begin{aligned}
f(x)= & \sqrt{x^{4}-3 x^{2}-6 x+13} \\
& -\sqrt{x^{4}-x^{2}+1}
\end{aligned}
$$

The maximum value is $\qquad$
$(1992$, National High School Mathematics Competition)
Solution: Transform the function, we get
$$
\begin{aligned}
f(x)= & \sqrt{(x-3)^{2}+\left(x^{2}-2\right)^{2}} \\
& -\sqrt{(x-0)^{2}+\left(x^{2}-1\right)^{2}} .
\end{aligned}
$$

It can be known that the geometric meaning of the function $y=f(x)$ is:
As shown in Figure 2, for a point $P(x, x^{2})$ on the parabola $y=x^{2}$, the difference in distances from $P$ to points $A(3, 2)$ and $B(0, 1)$. Find

its maximum value.
By $||PA|-|PB|| \leq |AB|$, we know that
when $P$ is at the extension of $AB$ at point $P_{0}$, $f(x)$ reaches its maximum value $|AB|$.
$$
\begin{aligned}
\therefore & {[f(x)]_{\max }} \\
& =\sqrt{(3-0)^{2}+(2-1)^{2}}=\sqrt{10 .}
\end{aligned}
$$