For any $y \in \mathbb{R}$, let $\cot^{-1}(y) \in (0, \pi)$ and $\tan^{-1}(y) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the sum of all solutions of the equation $\tan^{-1}\left(\frac{6y}{9-y^{2}}\right) + \cot^{-1}\left(\frac{9-y^{2}}{6y}\right) = \frac{2\pi}{3}$ for $0 < |y| < 3$, is equal to

For any y ∈ R, let cot⁻¹(y) ∈ (0, π) and tan⁻¹(y) ∈ (-π/2, π/2). Then the sum of all solutions of the equation tan⁻¹(6y/(9-y²)) + cot⁻¹((9-y²)/6y) = 2π/3 for 0 < |y| < 3, is equal to

$\cot^{-1}(y)\in(0,\pi),\quad \tan^{-1}(y)\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$

$\tan^{-1}\!\left(\dfrac{6y}{9-y^2}\right) + \cot^{-1}\!\left(\dfrac{9-y^2}{6y}\right) = \dfrac{2\pi}{3}$

$0 < |y| < 3$

$\text{find the sum of all solutions } y$

$1.\ u = \dfrac{6y}{9-y^2};\ \dfrac{9-y^2}{6y} = \dfrac{1}{u}$

$2.\ \text{two cases: } u>0,\ u<0$

$3.\ \text{solve for } y,\ \text{sum the admissible roots}$

$u = \dfrac{6y}{9-y^2}$

$\dfrac{9-y^2}{6y} = \dfrac{1}{u}$

$\tan^{-1}(u) + \cot^{-1}\!\left(\dfrac{1}{u}\right) = \dfrac{2\pi}{3}$

$u>0:\ \cot^{-1}\!\left(\tfrac{1}{u}\right) = \tan^{-1}(u)$

$u<0:\ \cot^{-1}\!\left(\tfrac{1}{u}\right) = \pi + \tan^{-1}(u)$

$\text{since } \cot^{-1}\in(0,\pi)$

$2\tan^{-1}(u) = \dfrac{2\pi}{3} \Rightarrow \tan^{-1}(u) = \dfrac{\pi}{3}$

$u = \sqrt{3}$

$\dfrac{6y}{9-y^2} = \sqrt{3} \Rightarrow y^2 + 2\sqrt{3}\,y - 9 = 0$

$y = \dfrac{-2\sqrt{3}\pm 4\sqrt{3}}{2} = \sqrt{3}\ \text{or}\ -3\sqrt{3}$

$0<|y|<3 \Rightarrow y = \sqrt{3}$

$\pi + 2\tan^{-1}(u) = \dfrac{2\pi}{3} \Rightarrow \tan^{-1}(u) = -\dfrac{\pi}{6}$

$u = -\dfrac{1}{\sqrt{3}}$

$\dfrac{6y}{9-y^2} = -\dfrac{1}{\sqrt{3}} \Rightarrow y^2 - 6\sqrt{3}\,y - 9 = 0$

$y = 3\sqrt{3} \pm 6$

$0<|y|<3 \Rightarrow y = 3\sqrt{3}-6$

$\text{sum} = \sqrt{3} + (3\sqrt{3}-6) = 4\sqrt{3}-6$

$\boxed{4\sqrt{3}-6}\quad \text{option (3)}$