9. (16 points) Given a positive number $p$ and the parabola $C: y^{2}=2 p x(p>0), A\left(\frac{p}{6}, 0\right)$ is a point on the axis of symmetry of the parabola $C$, $O$ is the vertex of the parabola $C$, and $M$ is any point on the parabola $C$. Find the maximum value of $\frac{|O M|}{|A M|}$.
9. Let $M(x, y)$, then $x \geqslant 0$, and $y^{2}=2 p x$.
$$
\left(\frac{|O M|}{|A M|}\right)^{2}=\frac{x^{2}+y^{2}}{\left(x-\frac{p}{6}\right)^{2}+y^{2}}=\frac{x^{2}+2 p x}{x^{2}-\frac{p}{3} x+\frac{p^{2}}{36}+2 p x}=\frac{x^{2}+2 p x}{x^{2}+\frac{5}{3} p x+\frac{p^{2}}{36}} \text {. }
$$

Let
$$
t=\frac{x^{2}+2 p x}{x^{2}+\frac{5}{3} p x+\frac{p^{2}}{36}},
$$

then
$$
(t-1) x^{2}+p\left(\frac{5}{3} t-2\right) x+\frac{p^{2}}{36} t=0 .
$$

When $t \neq 1$, since $x \in \mathbf{R}$, we have
$$
\Delta=p^{2}\left(\left(\frac{5}{3} t-2\right)^{2}-\frac{1}{9} t(t-1)\right)=p^{2}\left(\frac{24}{9} t^{2}+\frac{59}{9} t+4\right) \geqslant 0 .
$$

Solving this, we get $t \geqslant \frac{4}{3}$ or $t \leqslant \frac{9}{8}$.
When $t=\frac{9}{8}$, $x=\frac{p}{2}$. And for $x \geqslant 0$,
$$
\frac{x^{2}+2 p x}{x^{2}+\frac{5}{3} p x+\frac{p^{2}}{36}} \leqslant \frac{9}{8} \Leftrightarrow\left(x-\frac{p}{2}\right)^{2} \geqslant 0 .
$$

Thus, for all $x \geqslant 0, t \leqslant \frac{9}{8}$ (hence $t \geqslant \frac{4}{3}$ is impossible).
Therefore, the maximum value of $t$ is $\frac{9}{8}$, and thus, the maximum value of $\frac{|O M|}{|A M|}$ is $\frac{3 \sqrt{2}}{4}$.