The vertices of an equilateral triangle lie on the hyperbola $xy=1,$ and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?

$\textbf{(A)} \text{ 48} \qquad \textbf{(B)} \text{ 60} \qquad \textbf{(C)} \text{ 108} \qquad \textbf{(D)} \text{ 120} \qquad \textbf{(E)} \text{ 169}$
1. Let the three vertices of the equilateral triangle have coordinates \(\left(a, \frac{1}{a}\right), \left(b, \frac{1}{b}\right), \left(c, \frac{1}{c}\right)\). Assume that the centroid of the triangle is \((1, 1)\). The centroid of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
   \[
   \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)
   \]
   Therefore, we have:
   \[
   \frac{a + b + c}{3} = 1 \quad \text{and} \quad \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} = 1
   \]
   Simplifying these, we get:
   \[
   a + b + c = 3 \quad \text{and} \quad \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 3
   \]

2. Substituting \(c = 3 - (a + b)\) into the second equation, we get:
   \[
   \frac{1}{a} + \frac{1}{b} + \frac{1}{3 - (a + b)} = 3
   \]
   Simplifying, we have:
   \[
   \frac{a + b}{ab} + \frac{1}{3 - (a + b)} = 3
   \]

3. Since \(\left(a, \frac{1}{a}\right)\) and \(\left(b, \frac{1}{b}\right)\) are equidistant from \((1, 1)\), using the distance formula, we get:
   \[
   (a - 1)^2 + \left(\frac{1}{a} - 1\right)^2 = (b - 1)^2 + \left(\frac{1}{b} - 1\right)^2
   \]
   Simplifying, we have:
   \[
   (a - 1)^2 + \left(\frac{1 - a}{a}\right)^2 = (b - 1)^2 + \left(\frac{1 - b}{b}\right)^2
   \]
   \[
   (a - 1)^2 + \frac{(1 - a)^2}{a^2} = (b - 1)^2 + \frac{(1 - b)^2}{b^2}
   \]
   \[
   (a - 1)^2 \left(1 + \frac{1}{a^2}\right) = (b - 1)^2 \left(1 + \frac{1}{b^2}\right)
   \]

4. This implies:
   \[
   (a - 1)^2 \left(\frac{a^2 + 1}{a^2}\right) = (b - 1)^2 \left(\frac{b^2 + 1}{b^2}\right)
   \]
   \[
   (a - 1)^2 (a^2 + 1) = (b - 1)^2 (b^2 + 1)
   \]

5. We can divide by \(a - b\) because \(a\) and \(b\) are distinct. This implies:
   \[
   a + b - 2 = \frac{a + b - 2}{ab}
   \]
   Simplifying, we get:
   \[
   ab(a + b - 2) = a + b - 2
   \]
   Since \(a + b - 2 \neq 0\), we can divide both sides by \(a + b - 2\):
   \[
   ab = 1
   \]

6. Substituting \(ab = 1\) into the equation \(a + b + c = 3\), we get:
   \[
   a + b = 4
   \]
   Solving the quadratic equation \(t^2 - 4t + 1 = 0\), we get:
   \[
   t = 2 \pm \sqrt{3}
   \]
   Therefore, \(a\) and \(b\) are \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\).

7. The side length of the equilateral triangle can be found using the distance formula between \(\left(2 + \sqrt{3}, \frac{1}{2 + \sqrt{3}}\right)\) and \(\left(2 - \sqrt{3}, \frac{1}{2 - \sqrt{3}}\right)\):
   \[
   \text{Side length} = \sqrt{\left( (2 + \sqrt{3}) - (2 - \sqrt{3}) \right)^2 + \left( \frac{1}{2 + \sqrt{3}} - \frac{1}{2 - \sqrt{3}} \right)^2}
   \]
   Simplifying, we get:
   \[
   \text{Side length} = \sqrt{(2\sqrt{3})^2 + \left( \frac{(2 - \sqrt{3}) - (2 + \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \right)^2}
   \]
   \[
   \text{Side length} = \sqrt{12 + \left( \frac{-2\sqrt{3}}{1} \right)^2}
   \]
   \[
   \text{Side length} = \sqrt{12 + 12} = \sqrt{24} = 2\sqrt{6}
   \]

8. The area of an equilateral triangle with side length \(s\) is given by:
   \[
   \text{Area} = \frac{\sqrt{3}}{4} s^2
   \]
   Substituting \(s = 2\sqrt{6}\), we get:
   \[
   \text{Area} = \frac{\sqrt{3}}{4} (2\sqrt{6})^2 = \frac{\sqrt{3}}{4} \cdot 24 = 6\sqrt{3}
   \]

9. The square of the area is:
   \[
   (6\sqrt{3})^2 = 108
   \]

The final answer is \(\boxed{108}\)