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

Let us denote
$$\alpha=\tan^{-1}\!\left(\dfrac{6y}{\,9-y^{2}\,}\right),\qquad \beta=\cot^{-1}\!\left(\dfrac{9-y^{2}}{6y}\right)$$

According to the question, $$\alpha+\beta=\dfrac{2\pi}{3} \qquad\text{with}\qquad 0\lt |y|\lt 3$$

Because $$\beta$$ is a principal inverse-cotangent, $$\beta\in(0,\pi)$$, while $$\alpha\in\!\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$. Put
$$u=\dfrac{6y}{\,9-y^{2}\,} \Longrightarrow \alpha=\tan^{-1}(u)$$ Then $$\dfrac{9-y^{2}}{6y}=\dfrac{1}{u}\,,$$ so $$\beta=\cot^{-1}\!\left(\dfrac{1}{u}\right).$$

Relationship between $$\alpha$$ and $$\beta$$
For every real $$u\neq0$$:

• If $$u\gt0$$, then $$\alpha\in(0,\pi/2)$$ and $$\beta=\tan^{-1}(1/u)\in(0,\pi/2).$$  Thus $$\beta=\alpha$$ and $$\alpha+\beta=2\alpha.$$
• If $$u\lt0$$, then $$\alpha\in(-\pi/2,0)$$.  Now $$\tan\beta=\dfrac{1}{u}\lt0$$, hence $$\beta$$ lies in the second quadrant, $$\beta=\pi-\tan^{-1}(|1/u|).$$  Because $$|\alpha|=\tan^{-1}(|u|)$$, we obtain $$\beta=\pi-|\alpha|=\pi+\,\alpha$$ and therefore $$\alpha+\beta=\pi+2\alpha.$$

Hence the given equation splits into two cases.

Case 1: $$u\gt0$$
$$2\alpha=\dfrac{2\pi}{3}\Longrightarrow \alpha=\dfrac{\pi}{3}$$ $$u=\tan\alpha=\sqrt{3}$$

Substituting $$u=\sqrt{3}$$:
$$\dfrac{6y}{\,9-y^{2}\,}=\sqrt{3}$$ $$6y=\sqrt{3}\,(9-y^{2})$$ $$\sqrt{3}\,y^{2}+6y-9\sqrt{3}=0$$ Dividing by $$\sqrt{3}$$ gives $$y^{2}+2\sqrt{3}\,y-9=0.$$

Solving: $$y=\dfrac{-2\sqrt{3}\pm4\sqrt{3}}{2}$$ $$\Rightarrow\;y=\sqrt{3}\quad\text{or}\quad y=-3\sqrt{3}.$$ Because $$0\lt|y|\lt3$$, only $$y=\sqrt{3}$$ is admissible.

Case 2: $$u\lt0$$
$$\pi+2\alpha=\dfrac{2\pi}{3}\Longrightarrow 2\alpha=-\dfrac{\pi}{3}\Longrightarrow \alpha=-\dfrac{\pi}{6}$$ $$u=\tan\alpha=-\dfrac{1}{\sqrt{3}}$$

Substituting $$u=-\dfrac{1}{\sqrt{3}}$$:
$$\dfrac{6y}{\,9-y^{2}\,}=-\dfrac{1}{\sqrt{3}}$$ Multiply by $$\sqrt{3}$$:
$$6\sqrt{3}\,y=-(9-y^{2})$$ $$y^{2}-6\sqrt{3}\,y-9=0$$

Discriminant $$\Delta=(6\sqrt{3})^{2}+36=108+36=144,\; \sqrt{\Delta}=12.$$ Thus $$y=\dfrac{6\sqrt{3}\pm12}{2}=3\sqrt{3}\pm6.$$

We get $$y_1=3\sqrt{3}+6\,(|y_1|>3,\;\text{rejected}),$$ $$y_2=3\sqrt{3}-6\,(\approx-0.804,\;|y_2|<3,\;\text{accepted}).$$

All acceptable solutions: $$y=\sqrt{3},\;y=3\sqrt{3}-6.$$

Required sum:
$$\sqrt{3}+\bigl(3\sqrt{3}-6\bigr)=4\sqrt{3}-6.$$

Therefore the desired value is $$4\sqrt{3}-6$$, i.e. Option C.