3. (1997 National High School Competition Question) Let the two branches of the hyperbola $xy=1$ be $C_{1}$ and $C_{2}$, and the three vertices of the equilateral triangle $PQR$ lie on this hyperbola.
(1) Prove that $P$, $Q$, and $R$ 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 $Q$ and $R$.
3. (1) Use proof by contradiction. Assume that the three vertices $P, Q, R$ of the equilateral $\triangle PQR$ are all on the same branch (such as $C_{1}$), with coordinates $P(x_{1}, y_{1}), Q(x_{2}, y_{2}), R(x_{3}, y_{3}) (0 < y_{1} < y_{2} > y_{3})$. Thus, $PQ^{2} + QR^{2} - PR^{2} = \left[(x_{1} - x_{2})^{2} + (x_{2} - x_{3})^{2} - (x_{1} - x_{3})^{2}\right] + \left[(y_{1} - y_{2})^{2} + (y_{2} - y_{3})^{2} - (y_{1} - y_{3})^{2}\right] = 2(x_{2} - x_{1})(x_{2} - x_{3}) + 2(y_{2} - y_{1})(y_{2} - y_{3})x_{1} > 0\right)$, the altitude from $P$ to $QR$ has the equation $y + 1 = x_{1} x_{2}(x + 1)$, which must pass through the midpoint of $QR \left(\frac{x_{1} + x_{2}}{2}, \frac{1}{2}\left(\frac{1}{x_{1}} + \frac{1}{x_{2}}\right)\right)$, so $\frac{1}{2}\left(\frac{1}{x_{1}} + \frac{1}{x_{2}}\right) + 1 = x_{1} x_{2}\left(\frac{x_{1} + x_{2}}{2} + 1\right)$, which simplifies to $\left(1 - x_{1} x_{2}\right)\left[(x_{1} + x_{2})(1 + x_{1} x_{2}) + 2 x_{1} x_{2}\right] = 0$, hence $x_{1} x_{2} = 1$, so the coordinates of $Q$ are $Q\left(\frac{1}{x_{2}}, x_{2}\right)$, and thus $Q\left(\frac{1}{x_{2}}, x_{2}\right)$ and $R\left(x_{2}, \frac{1}{x_{2}}\right)$ are symmetric about the line $y = x$, therefore the angles between $PQ$ and $PR$ and the line $y = x$ are both $30^{\circ}$, so the inclinations of $PQ$ and $PR$ are $75^{\circ}$ and $30^{\circ}$ respectively. At this point, the equation of $PQ$ is $y + 1 = \tan 75^{\circ}(x + 1)$, i.e., $y + 1 = (2 + \sqrt{3})(x + 1)$, and solving this with $y = \frac{1}{x}$ gives the coordinates of $Q$ as $(2 - \sqrt{3}, 2 + \sqrt{3})$. By symmetry, the coordinates of $R$ are $(2 + \sqrt{3}, 2 - \sqrt{3})$.