Question:

**Step 1: Analyze the properties of inverse trigonometric functions** We know that $\cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x)$ if $x > 0$, and $\cot^{-1}\left(\frac{1}{x}\right) = \pi + \tan^{-1}(x)$ if $x < 0$. Let $x = \frac{6y}{9-y^2}$. Since $0 < |y| < 3$, we have $9-y^2 > 0$. Thus, the sign of $x$ is the same as the sign of $y$. **Step 2: Case 1: y is positive** If $0 < y < 3$, then $x > 0$. Therefore, $\cot^{-1}(1/x) = \tan^{-1}(x)$. The equation becomes $2\tan^{-1}\left(\frac{6y}{9-y^2}\right) = \frac{2\pi}{3}$, or $\tan^{-1}\left(\frac{6y}{9-y^2}\right) = \frac{\pi}{3}$.\nThis gives $\frac{6y}{9-y^2} = \sqrt{3}$. Rearranging: $\sqrt{3}y^2 + 6y - 9\sqrt{3} = 0 \implies y^2 + 2\sqrt{3}y - 9 = 0$.\nSolving for $y$: $y = \frac{-2\sqrt{3} \pm \sqrt{12 + 36}}{2} = -\sqrt{3} \pm 2\sqrt{3}$.\nSince $y > 0$, the valid solution is $y_1 = \sqrt{3}$. **Step 3: Case 2: y is negative** If $-3 < y < 0$, then $x < 0$. Therefore, $\cot^{-1}(1/x) = \pi + \tan^{-1}(x)$. The equation becomes $2\tan^{-1}\left(\frac{6y}{9-y^2}\right) + \pi = \frac{2\pi}{3}$, or $\tan^{-1}\left(\frac{6y}{9-y^2}\right) = -\frac{\pi}{6}$.\nThis gives $\frac{6y}{9-y^2} = -\frac{1}{\sqrt{3}}$. Rearranging: $9-y^2 = -6\sqrt{3}y \implies y^2 - 6\sqrt{3}y - 9 = 0$.\nSolving for $y$: $y = \frac{6\sqrt{3} \pm \sqrt{108 + 36}}{2} = 3\sqrt{3} \pm 6$.\nSince $y < 0$, the valid solution is $y_2 = 3\sqrt{3} - 6$. **Step 4: Calculate the sum of solutions** The solutions are $y_1 = \sqrt{3}$ and $y_2 = 3\sqrt{3} - 6$. Their sum is $\sqrt{3} + 3\sqrt{3} - 6 = 4\sqrt{3} - 6$.