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

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

$\mathrm{O}<\left|\mathrm{y}\right|<\sqrt{3}$

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

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

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

$\begin{array}{l}{\tan }^{-1}\left(\frac{6y}{9-y^2}\right)+\pi +{\tan }^{-1}\left(\frac{6y}{9-y^2}\right)=\frac{2\pi }{3}\\ \Rightarrow 2{\tan }^{-1}\left(\frac{6y}{9-y^2}\right)=-\frac{\pi }{3}\Rightarrow {\tan }^{-1}\left(\frac{6y}{9-y^2}\right)=-\frac{\pi }{6}\\ \Rightarrow \frac{6y}{9-y^2}=-\frac{1}{\sqrt{3}}\\ \Rightarrow 6\sqrt{3}y=-9+y^2\\ \Rightarrow y^2-6\sqrt{3}y-9=0\\ \Rightarrow y=\frac{6\sqrt{3}\pm \sqrt{108+36}}{2}=\frac{6\sqrt{3}\pm 12}{2}=3\sqrt{3}\pm 6\end{array}$

$$\begin{gathered} \text{ as }y\in (-3.0)\therefore y=3\sqrt{3}-6 \\ \therefore \text{ Sum of solutions }=\sqrt{3}+(3\sqrt{3}-6) \\ =4\sqrt{3}-6 \end{gathered}$$