Concept :

$$ \cot ^{-1} x=\left\{\begin{aligned} \pi+\tan ^{-1} \frac{1}{x}; & x<0 \\ \tan ^{-1} \frac{1}{x} & ; x>0 \end{aligned}\right. $$

Solution : Given, $0 < |y| < 3$

$$ \Rightarrow $$ $$ y \in(-3,3)- $$ {0}

$$9 - {y^2}$$ is always positive when $$ y \in(-3,3)- $$ {0}

And $6y$ is positive when $$ y \in(0,3) $$

And $6y$ is negative when $$ y \in(-3,0) $$

$$ \therefore $$ In overall, $$ \frac{6 y}{9-y^2} > 0 $$ when $$ y \in(0,3) $$

And $$ \frac{6 y}{9-y^2} < 0 $$ when $$ y \in(-3,0) $$

Case - 1 :

When $-3 < y < 0$

$$ \begin{aligned} & \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\pi+\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \\\\ & \Rightarrow 2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{-\pi}{3} \\\\ & \Rightarrow \frac{6 y}{9-y^2}=\frac{-1}{\sqrt{3}} \\\\ & \Rightarrow y^2-6 \sqrt{3} y-9=0 \\\\ & \Rightarrow y^2-6 \sqrt{3} y-9=0 \end{aligned} $$

$$ \begin{aligned} & \Rightarrow y=\frac{6 \sqrt{3} \pm \sqrt{108+36}}{2}=\frac{6 \sqrt{3} \pm 12}{2}=3 \sqrt{3} \pm 6 \\\\ & \text { as } y \in(-3.0) \therefore y=3 \sqrt{3}-6 \end{aligned} $$

Case-2 : When $0 < y < 3$

$$ \begin{aligned} & \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \\\\ & \Rightarrow 2 \tan ^{-1}\left(\frac{6 y}{9-y^2}\right)=\frac{2 \pi}{3} \\\\ & \Rightarrow \frac{6 y}{9-y^2}=\sqrt{3} \end{aligned} $$

$$ \begin{aligned} & \Rightarrow 6 y=9 \sqrt{3}-\sqrt{3} y^2 \\\\ & \Rightarrow \sqrt{3} y^2+6 y-9 \sqrt{3}=0 \\\\ & \Rightarrow \sqrt{3} y^2+9 y-3 y-9 \sqrt{3}=0 \\\\ & \Rightarrow \sqrt{3} y(y+3 \sqrt{3})-3(y+3 \sqrt{3})=0 \\\\ & \Rightarrow(y+3 \sqrt{3})-(\sqrt{3} y-3)=0 \end{aligned} $$

$$ y \neq-3 \sqrt{3} \quad \therefore y=\sqrt{3} \text { as } y \in(0,3) $$

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