Example 1 (1997 National High School Competition Question) Let the two branches of the hyperbola $xy=1$ be $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ as shown in Figure 7-1. The vertices of the equilateral triangle $\mathrm{PQR}$ are located on this hyperbola.
(1) Prove that $P, Q, R$ cannot all be 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 vertices $Q$ and $R$.
(1) Assume the three vertices $P, Q, R$ of the equilateral $\triangle PQR$ are on the same branch of $C_{1}$, 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, let $0 < y_{1} < y_{2} < y_{3} < 0$.
Thus, $P Q^{2}+Q R^{2}-P R^{2}$
$$
\begin{aligned}
= & {\left[\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}-\left(x_{1}-x_{3}\right)^{2}\right]+\left[\left(y_{1}-y_{2}\right)^{2}+\right.} \\
& \left.\left(y_{2}-y_{3}\right)^{2}-\left(y_{1}-y_{3}\right)^{2}\right] \\
= & \left(2 x_{2}^{2}-2 x_{1} x_{2}-2 x_{2} x_{3}+2 x_{1} x_{3}\right)+\left(2 y_{2}^{2}-2 y_{1} y_{2}-\right. \\
& \left.2 y_{2} y_{3}+2 y_{1} y_{3}\right) \\
= & 2\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)+2\left(y_{2}-y_{1}\right)\left(y_{2}-y_{3}\right) < 0
\end{aligned}
$$
Since $x_{1} < 0, x_{2} > 0$, 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 point $Q$ are $\left(\frac{1}{x_{2}}, x_{1}\right)$, and the coordinates of point $R$ are $\left(x_{2}, \frac{1}{x_{2}}\right)$, which means $Q$ and $R$ are symmetric about the line $y = x$.
The lines $PQ$ and $PR$ are two mutually symmetric lines passing through point $P$ and intersecting the line $y = x$ at an angle of $30^{\circ}$. It is easy to see that their slopes are $75^{\circ}$ and $15^{\circ}$, respectively. Let the slope of $PQ$ be $75^{\circ}$, then its equation is $y + 1 = \tan 75^{\circ} \cdot (x + 1)$, i.e., $y + 1 = (2 + \sqrt{3})(x + 1)$. Substituting this into the hyperbola equation $xy = 1$, we solve to get the coordinates of point $Q$ as $(2 - \sqrt{3}, 2 + \sqrt{3})$. By symmetry, the coordinates of point $R$ are $(2 + \sqrt{3}, 2 - \sqrt{3})$.