Example 6 Find the minimum value of the function
$$
\begin{array}{c}
f(x)=\sqrt{9(x-3)^{2}+\left(x^{2}-12\right)^{2}}+ \\
\sqrt{9 x^{2}+\left(x^{2}-\frac{9}{4}\right)^{2}}
\end{array}
$$
Given that
$$
\begin{array}{c}
\frac{1}{3} f(x)=\sqrt{(x-3)^{2}+\left(\frac{x^{2}}{3}-4\right)^{2}}+ \\
\sqrt{x^{2}+\left(\frac{x^{2}}{3}-\frac{3}{4}\right)^{2}},
\end{array}
$$

we can let $y=\frac{x^{2}}{3}$, which is the equation of a parabola, with the focus at $F\left(0, \frac{3}{4}\right)$ and the directrix equation as $y=-\frac{3}{4}$.
As shown in Figure 2, let point $A(3,4)$.
Then equation (1) can be rewritten as
$$
\begin{array}{l}
\frac{1}{3} f(x) \\
=\sqrt{(x-3)^{2}+(y-4)^{2}}+\sqrt{x^{2}+\left(y-\frac{3}{4}\right)^{2}},
\end{array}
$$

which represents the sum of the distances from a point $M(x, y)$ on the parabola to point $A$ and to the focus $F$: $\frac{1}{3} f = M A + M F$.
Note that point $A$ is above the parabola.
Since the distance from point $M$ to the focus is equal to its distance to the directrix $M F = M H$, the value of $M A + M H$ is minimized when point $M$ moves to $M_{1}$, i.e., when $M$ is on the perpendicular line $A H_{1}$, which is
$$
\begin{array}{l}
A M_{1} + M H_{1} = A H_{1} = 4 + \frac{3}{4} = \frac{19}{4} \\
\Rightarrow \frac{1}{3} f \geqslant \frac{19}{4} \Rightarrow f \geqslant \frac{57}{4} .
\end{array}
$$