14. Let the two branches of the hyperbola $xy=1$ be $C_{1}$ and $C_{2}$ (as shown in the figure), and the 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 the vertices $Q$ and $R$.
(1) Without loss of generality, let $P\left(a, \frac{1}{a}\right), Q\left(b, \frac{1}{b}\right), R\left(c, \frac{1}{c}\right)$, where $0 < a < b < c$. If $P, Q, R$ are all on the same branch of the hyperbola, then $\angle PQR > 90^{\circ}$, which contradicts the fact that $\triangle PQR$ is an equilateral triangle.
Therefore, $P, Q, R$ cannot all be on the same branch of the hyperbola.
(2) By the symmetry of the hyperbola, points $Q, R$ are symmetric with respect to the line
$y=x$. Let $Q\left(\frac{1}{c}, c\right), R\left(c, \frac{1}{c}\right)(c>1)$,
then $|QR|=\sqrt{2}\left(c-\frac{1}{c}\right), QR: x+y-c-\frac{1}{c}=0$.
Thus, the distance from point $P$ to the line $QR$ is
$$
\begin{array}{l}
d=\frac{2+c+\frac{1}{c}}{\sqrt{2}}=\frac{\sqrt{3}}{2} \cdot \sqrt{2}\left(c-\frac{1}{c}\right) \\
\Rightarrow 2+c+\frac{1}{c}=\sqrt{3}\left(c-\frac{1}{c}\right) \Rightarrow (\sqrt{3}-1) c^{2}-2 c-(\sqrt{3}+1) \\
=[(\sqrt{3}-1) c-(\sqrt{3}+1)](c+1)=0 \Rightarrow c=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3} .
\end{array}
$$

Thus, we get $Q(2-\sqrt{3}, 2+\sqrt{3}), R(2+\sqrt{3}, 2-\sqrt{3})$.