9. It is easy to see that the center of symmetry of the graph of $f(x)=x^{3}-4 x^{2}+4 x$ is $\left(\frac{4}{3}, \frac{16}{27}\right)$.

Translating the center of symmetry of $f(x)=x^{3}-4 x^{2}+4$ to the point $(0,0)$ yields the function
$$
y=f\left(x+\frac{4}{3}\right)-\frac{16}{27}=x^{3}-\frac{4}{3} x .
$$

The equation of the tangent line passing through point $P_{1}$ is
$$
y-\left(x_{1}^{3}-\frac{4}{3} x_{1}\right)=\left(3 x_{1}^{2}-\frac{4}{3}\right)\left(x-x_{1}\right) \text {, }
$$

which simplifies to $y=\left(3 x_{1}^{2}-\frac{4}{3}\right) x-2 x_{1}^{3}$.
By setting $\left(3 x_{1}^{2}-\frac{4}{3}\right) x-2 x_{1}^{3}=x^{3}-\frac{4}{3} x$, we get $x_{2}=-2 x_{1}$.
Then, $S_{1}=\left|\int_{x_{1}}^{-2 x_{1}}\left[\left(3 x_{1}^{2}-\frac{4}{3}\right) x-2 x_{1}^{3}-\left(x^{3}-\frac{4}{3} x\right)\right] \mathrm{d} x\right|$ $=\frac{27}{4} x_{1}^{4}$.
Similarly, by substituting $x_{2}$ for $x_{1}$, we get $x_{3}=-2 x_{2}$. Thus, $S_{2}=\frac{27}{4} x_{2}^{4}=\frac{27}{4}\left(-2 x_{1}\right)^{4}=\frac{27 \times 16}{4} x_{1}^{4}$.
Therefore, $\frac{S_{1}}{S_{2}}=\frac{1}{16}$.
9. (16 points) As shown in Figure 1, given the function
$$
f(x)=x^{3}-4 x^{2}+4 x \text {, }
$$

its graph is denoted as curve $C$. If for any $x_{1} \neq \frac{4}{3}$, curve $C$ intersects with the tangent line at point $P_{1}\left(x_{1}, f\left(x_{1}\right)\right)$ at another point $P_{2}\left(x_{2}, f\left(x_{2}\right)\right)$, and curve $C$ intersects with the tangent line at point $P_{2}$ at another point $P_{3}\left(x_{3}, f\left(x_{3}\right)\right)$, the areas of the closed figures formed by line segments $P_{1} P_{2}$, $P_{2} P_{3}$, and curve $C$ are denoted as $S_{1}, S_{2}$, respectively. Find the value of $\frac{S_{1}}{S_{2}}$.