For any y ∈ ℝ , let cot - 1 y ∈ 0 , π and tan - 1 y ∈ - π 2 , π 2 .…

Given,

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

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

Now taking, Case-l:

When $\frac{6y}{9-y^2}>0\Rightarrow y>0$

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

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

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

$\Rightarrow \frac{6y}{9-y^2}=\sqrt{3}$

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

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

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

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

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

Now taking, Case-II: 

When $\frac{6y}{9-y^2}<0\Rightarrow y<0$

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

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

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

$\Rightarrow \frac{6y}{9-y^2}=-\frac{1}{\sqrt{3}}$

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

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

$\Rightarrow y=\frac{6\sqrt{3}\pm \sqrt{108+36}}{2}=\frac{6\sqrt{3}\pm 12}{2}=3\sqrt{3}\pm 6$

$\text{ As }y\in (-3.0), \text{so} y=3\sqrt{3}-6$

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