Write $u=\frac{6y}{9-y^2}$. Then the second term equals ${\cot }^{-1}(1/u)$, which is ${\tan }^{-1}(u)$ if $u>0$ and $\pi +{\tan }^{-1}(u)$ if $u<0$.

Case 1: $y\in (0,3)$. Here $u>0$, so the equation becomes $2{\tan }^{-1}u=\frac{2\pi }{3}$, giving $u=\tan \frac{\pi }{3}=\sqrt{3}$. Thus $\sqrt{3}{y}^2+6y-9\sqrt{3}=0$, whose roots are $y=\sqrt{3}$ (accepted) and $y=-3\sqrt{3}$ (rejected).

Case 2: $y\in (-3,0)$. Here $u<0$, so the equation becomes $2{\tan }^{-1}u+\pi =\frac{2\pi }{3}$, giving ${\tan }^{-1}u=-\frac{\pi }{6}$ and $u=-\frac{1}{\sqrt{3}}$. Thus $y^2-6\sqrt{3}y-9=0$, whose roots are $3\sqrt{3}\pm 6$. Only $y=3\sqrt{3}-6\approx -0.8$ lies in $(-3,0)$.

Sum of solutions $=\sqrt{3}+3\sqrt{3}-6=4\sqrt{3}-6.$