Let the two branches of the hyperbola $xy=1$ be $C_{1}$ and $C_{2}$ (as shown in the figure), and the three vertices of the equilateral triangle $PQR$ 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) Proof by contradiction. Assume that the three vertices $P, Q, R$ of the equilateral $\triangle PQR$ are on the same branch, such as $C_{1}$, with coordinates $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right), \left(x_{3}, y_{3}\right)$, respectively. Without loss of generality, assume $0 < y_{1} < y_{2} < y_{3} < 0$. Thus,
$$
\begin{aligned}
& |P Q|^{2}+|Q R|^{2}-|P R|^{2} \\
= & {\left[\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}-\left(x_{1}-x_{3}\right)^{2}\right] } \\
& +\left[\left(y_{1}-y_{2}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}-\left(y_{1}-y_{3}\right)^{2}\right] \\
= & \left(2 x_{2}^{2}-2 x_{1} x_{2}-2 x_{2} x_{3}+2 x_{1} x_{3}\right) \\
& +\left(2 y_{2}^{2}-2 y_{1} y_{2}-2 y_{2} y_{3}+2 y_{1} y_{3}\right) \\
= & 2\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)+2\left(y_{2}-y_{1}\right)\left(y_{2}-y_{3}\right) < 0.
\end{aligned}
$$
Since $x_{1} x_{2} x_{3} = 1$ and $x_{1} x_{2} x_{3} > 0, x_{2} > 0$, the expression in the square brackets is clearly greater than 0, so $1 - x_{1} x_{2} = 0$. Hence, $x_{1} x_{2} = 1$. Therefore, the coordinates of $Q$ are $\left(\frac{1}{x_{2}}, x_{2}\right)$, and the coordinates of $R$ are $\left(x_{2}, \frac{1}{x_{2}}\right)$, which means $Q$ and $R$ are symmetric with respect to the line $y = x$. The lines $PQ$ and $PR$ are two mutually symmetric lines passing through point $P$ and forming a $30^{\circ}$ angle with $y = x$. It is easy to see that their slopes are $75^{\circ}$ and $15^{\circ}$, respectively. Without loss of generality, assume the slope of $PQ$ is $75^{\circ}$, so its equation is $y + 1 = \tan 75^{\circ} \cdot (x + 1)$. That is, $y + 1 = (2 + \sqrt{3})(x + 1)$. Substituting this into the hyperbola equation $xy = 1$, we find that the coordinates of $Q$ are $(2 - \sqrt{3}, 2 + \sqrt{3})$, and by symmetry, the coordinates of $R$ are $(2 + \sqrt{3}, 2 - \sqrt{3})$.