Let the two branches of the hyperbola $x y=1$ be $C_{1}$ and $C_{2}$ (as shown in the figure), and the three vertices of the equilateral triangle $P Q R$ lie on this hyperbola.
(1) Prove that $P, Q, R$ cannot all lie on the same branch of the hyperbola;
(2) Suppose $P(-1,-1)$ is on $C_{2}$, and $Q, R$ are on $C_{1}$. Find the coordinates of the vertices $Q, R$.
(1) By contradiction.
Assume the three vertices $P, Q, R$ of the equilateral $\triangle PQR$ are all on the same branch, say $C_1$, with coordinates $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}+y_{3}\right)$, respectively. Without loss of generality, assume $0 < y_{1} < y_{2} < y_{3}$. Thus,
\[
\begin{aligned}
& P Q^{2}+Q R^{2}-P R^{2} \\
= & \left[\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}-\left(x_{1}-x_{3}\right)^{2}\right]+\left[\left(y_{1}-y_{2}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}\right. \\
& \left.-\left(y_{1}-y_{3}\right)^{2}\right] \\
= & \left(2 x_{2}^{2}-2 x_{1} x_{2}-2 x_{2} x_{3}+2 x_{1} x_{3}\right)+\left(2 y_{2}^{2}-2 y_{1} y_{2}-2 y_{2} y_{3}+2 y_{1} y_{3}\right) \\
= & 2\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)+2\left(y_{2}-y_{1}\right)\left(y_{2}-y_{3}\right) > 0, \quad \text{since} \quad x_{2}>0, \quad y_{2}>0, \quad x_{2}>0, \quad y_{2}>0.
\end{aligned}
\]
The expression in the brackets is clearly greater than 0, so $1 - x_{1} x_{2} = 0$. Hence, $x_{1} x_{2} = 1$.
Thus, the coordinates of $Q$ are $\left(\frac{1}{x_{2}}, x_{2}\right)$, and the coordinates of $R$ are $\left(x_{2}, \frac{1}{x_{2}}\right)$, indicating that $Q$ and $R$ are symmetric with respect to the line $y = x$.
The lines $PQ$ and $PR$ are two mutually symmetric lines passing through $P$ and intersecting $y = x$ at an angle of $30^\circ$. Their slopes are $75^\circ$ and $15^\circ$, respectively. Assume the slope of $PQ$ is $75^\circ$, so its equation is
\[
y + 1 = \tan 75^\circ \cdot (x + 1),
\]
which simplifies to
\[
y + 1 = (2 + \sqrt{3})(x + 1).
\]
Substituting this into the hyperbola equation $xy = 1$, we find the coordinates of $Q$ to be $(2 - \sqrt{3}, 2 + \sqrt{3})$. By symmetry, the coordinates of $R$ are $(2 + \sqrt{3}, 2 - \sqrt{3})$.