9. (16 points) Given that the three vertices of the equilateral triangle $\triangle ABC$ lie on the parabola $y=x^{2}$. Try to find the equation of the locus of the center of the equilateral triangle $\triangle ABC$.


9. Let $l_{A B}: y=k x+b$.
(1)
By symmetry, we first assume $k \geqslant 0$.
Combining equation (1) with $y=x^{2}$, we get
$$
x^{2}-k x-b=0.
$$

Then $x_{A}+x_{B}=k, x_{A} x_{B}=-b, \Delta=k^{2}+4 b>0$.
Thus, the midpoint $M\left(\frac{k}{2}, \frac{k^{2}}{2}+b\right)$ of $A B$, and
$$
|A B|=\sqrt{1+k^{2}}\left|x_{A}-x_{B}\right|=\sqrt{\left(1+k^{2}\right) \Delta}.
$$

Then $l_{C M}: y=-\frac{1}{k}\left(x-\frac{k}{2}\right)+\frac{k^{2}}{2}+b$, i.e.,
$$
y=-\frac{x}{k}+\frac{k^{2}+2+\Delta}{4}.
$$

By $|C M|=\frac{\sqrt{3}}{2}|A B|$
$$
\begin{array}{l}
\Rightarrow \sqrt{1+\frac{1}{k^{2}}}\left|x_{c}-\frac{k}{2}\right|=\frac{\sqrt{3}}{2} \sqrt{\left(1+k^{2}\right) \Delta} \\
\Rightarrow\left|x_{c}-\frac{k}{2}\right|=\frac{k}{2} \sqrt{3 \Delta} \\
\Rightarrow x_{c}=\frac{k}{2}(1 \pm \sqrt{3 \Delta})=\frac{k}{2}(1+\varepsilon \sqrt{3 \Delta}),
\end{array}
$$

where, $\varepsilon \in\{1,1\}$.
Combining equation (2) with $y=x^{2}$, we get
$$
\begin{aligned}
x_{c}^{2} & =-\frac{x_{c}}{k}+\frac{k^{2}+2+\Delta}{4} \\
& \Rightarrow \frac{k^{2}}{4}(1+3 \Delta+2 \varepsilon \sqrt{3 \Delta})=-\frac{\varepsilon \sqrt{3 \Delta}}{2}+\frac{k^{2}+\Delta}{4} \\
& \Rightarrow 2 \varepsilon\left(1+\dot{k}^{2}\right) \sqrt{3 \Delta}=\left(1-3 k^{2}\right) \Delta \\
& \Rightarrow \sqrt{\Delta}=\frac{2 \sqrt{3}\left(1+k^{2}\right)}{1-3 k^{2}} \varepsilon,
\end{aligned}
$$

where, $k^{2} \neq \frac{1}{3}$, and
$$
\varepsilon=\operatorname{sgn}\left(1-3 k^{2}\right)=\left\{\begin{array}{ll}
1, & 1-3 k^{2}>0 ; \\
-1, & 1-3 k^{2}<0 .
\end{array}\right.
$$

Thus, $\Delta=12\left(\frac{k^{2}+1}{1-3 k^{2}}\right)^{2}$.
Hence $x_{c}=\frac{k}{2}(1+\varepsilon \sqrt{3 \Delta})$
$$
=\frac{k}{2}\left[1+\frac{6\left(k^{2}+1\right)}{1-3 k^{2}}\right]=\frac{k}{2} \cdot \frac{3 k^{2}+7}{1-3 k^{2}}.
$$

Let the centroid of $\triangle A B C$ be $O(x, y)$. Then
$$
\begin{aligned}
x & =\frac{1}{3}\left(2 x_{M}+x_{C}\right) \\
& =\frac{k}{6}\left(2+\frac{3 k^{2}+7}{1-3 k^{2}}\right)=\frac{k}{2} \cdot \frac{3-k^{2}}{1-3 k^{2}}, \\
y & =-\frac{x}{k}+\frac{k^{2}+2+\Delta}{4} \\
& =\frac{9 k^{6}+18 k^{4}+33 k^{2}+8}{4\left(1-3 k^{2}\right)^{2}},
\end{aligned}
$$

where, the parameter $k \neq \pm \frac{\sqrt{3}}{3}$.