For any y ∈ ℝ , let ^ - 1 (y) ∈ (0, π) and ^ - 1 (y) ∈ (- (π)/(2),...

Given,

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

And $0 < | y | < 3 ⇒ y \in \left(- 3, 3\right) - \left\{0\right\}$

Now taking, Case-l:

When $\dfrac{6y}{9 - {y}^{2}} > 0 ⇒ y > 0$

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

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

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

$⇒ \dfrac{6y}{9 - {y}^{2}} = \sqrt{3}$

$⇒ 6y = 9\sqrt{3} - \sqrt{3}{y}^{2}$

$⇒ \sqrt{3}{y}^{2} + 6y - 9\sqrt{3} = 0$

$⇒ \sqrt{3}{y}^{2} + 9y - 3y - 9\sqrt{3} = 0$

$⇒ \left(y + 3\sqrt{3}\right)\left(\sqrt{3}y - 3\right) = 0$

$y \ne - 3\sqrt{3} ∴ y = \sqrt{3}$ as $y \in ( 0, 3 )$

Now taking, Case-II:

When $\dfrac{6y}{9 - {y}^{2}} < 0 ⇒ y < 0$

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

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

$⇒ \tan^{- 1}\left(\dfrac{6y}{9 - {y}^{2}}\right) = - \dfrac{\pi}{6}$

$⇒ \dfrac{6y}{9 - {y}^{2}} = - \dfrac{1}{\sqrt{3}}$

$⇒ 6\sqrt{3}y = - 9 + {y}^{2}$

$⇒ {y}^{2} - 6\sqrt{3}y - 9 = 0$

$⇒ y = \dfrac{6\sqrt{3} \pm \sqrt{108 + 36}}{2} = \dfrac{6\sqrt{3} \pm 12}{2} = 3\sqrt{3} \pm 6$

$\text{As}y \in ( - 3 .0 ) , \text{so}y = 3\sqrt{3} - 6$

$∴ \text{Sum of solutions} = \sqrt{3} + ( 3\sqrt{3} - 6 ) = 4\sqrt{3} - 6$