2. Given that the equilateral triangle $\triangle ABC$ is inscribed in the parabola $x=y^{2}$, and the centroid $P$ of $\triangle ABC$ lies on the hyperbola $x y=1$. Find the coordinates of point $P$.
2. Let $A\left(y_{1}^{2}, y_{1}\right), B\left(y_{2}^{2}, y_{2}\right), C\left(y_{3}^{2}, y_{3}\right)$, and the midpoints of $A B$ and $B C$ be $F$ and $D$, respectively. Then $\triangle A B C$ is an equilateral triangle if and only if $y_{1}, y_{2}, y_{3}$ are distinct, and $A D \perp B C, C F \perp A B$.
Thus, $D\left(\frac{y_{2}^{2}+y_{3}^{2}}{2}, \frac{y_{2}+y_{3}}{2}\right)$.
From $A D \perp B C, y_{2} \neq y_{3}$, we get
$$
\begin{array}{l}
\left(y_{2}^{2}-y_{3}^{2}\right)\left(\frac{y_{2}^{2}+y_{3}^{2}}{2}-y_{1}^{2}\right)+\left(y_{2}-y_{3}\right)\left(\frac{y_{2}+y_{3}}{2}-y_{1}\right)=0 \\
\Rightarrow\left(y_{2}+y_{3}\right)\left(y_{2}^{2}+y_{3}^{2}-2 y_{1}^{2}\right)+\left(y_{2}+y_{3}-2 y_{1}\right) \\
=0 .
\end{array}
$$

Similarly, from $C F \perp A B, y_{1} \neq y_{2}$, we get
$$
\left(y_{1}+y_{2}\right)\left(y_{1}^{2}+y_{2}^{2}-2 y_{3}^{2}\right)+\left(y_{1}+y_{2}-2 y_{3}\right)=0 \text {. }
$$
$$
\begin{array}{l}
\text { (1)-(2) gives } \\
\left(y_{3}^{3}-y_{1}^{3}\right)+3 y_{2}\left(y_{3}^{2}-y_{1}^{2}\right)+ \\
\left(y_{3}-y_{1}\right)\left(y_{2}^{2}+2 y_{1} y_{3}\right)+3\left(y_{3}-y_{1}\right)=0 \\
\Rightarrow\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)+3\left(y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1}\right)+3 \\
\quad=0 .
\end{array}
$$

Let $P(x, y)$. Then
$$
x=\frac{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}{3}, y=\frac{y_{1}+y_{2}+y_{3}}{3} \text {. }
$$

Thus, from equation (3) we get
$$
3 x+3 \times \frac{9 y^{2}-3 x}{2}+3=0 \Rightarrow 9 y^{2}-x+2=0 \text {. }
$$

Also, from $x y=1 \Rightarrow y=\frac{1}{x}$, substituting into the above equation and simplifying, we get
$$
\begin{array}{l}
x^{3}-2 x^{2}-9=0 \\
\Rightarrow(x-3)\left(x^{2}+x+3\right)=0 \Rightarrow x=3 .
\end{array}
$$

Therefore, $P\left(3, \frac{1}{3}\right)$.