For any y ∈ R, let cot⁻¹(y) ∈ (0, π) and tan⁻¹(y) ∈ [−π/2, π/2]. 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:

The correct option is $4\sqrt{3}-6$.

Step 1: Analyze the given condition for $y$
We are given that $0<\left|y\right|<3$. This means that $y\ne 0$ and $-3<y<3$.

Let us analyze the expression $x=\frac{6y}{9-y^2}$.
Since $-3<y<3$, we have $y^2<9$, which implies $9-y^2>0$.

Case 1: If $0<y<3$, then $6y>0$ and $9-y^2>0$, so $x>0$.
Case 2: If $-3<y<0$, then $6y<0$ and $9-y^2>0$, so $x<0$.

Step 2: Simplify the equation based on the properties of inverse trigonometric functions
Recall the relationship between ${cot}^{-1}\left(\frac{1}{x}\right)$ and ${tan}^{-1}\left(x\right)$:

${cot}^{-1}\left(\frac{1}{x}\right)={tan}^{-1}\left(x\right)$ when $x>0$
${cot}^{-1}\left(\frac{1}{x}\right)=\pi +{tan}^{-1}\left(x\right)$ when $x<0$

Here, the given equation is:

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

Let us evaluate both cases for $y$:

Case 1: $0<y<3$ (which means $x>0$)
Since $x>0$, ${cot}^{-1}\left(\frac{1}{x}\right)={tan}^{-1}\left(x\right)$.
Substituting this into the equation gives:

${tan}^{-1}\left(x\right)+{tan}^{-1}\left(x\right)=\frac{2\pi }{3}$
$2{tan}^{-1}\left(x\right)=\frac{2\pi }{3}$
${tan}^{-1}\left(x\right)=\frac{\pi }{3}$
$x=tan\left(\frac{\pi }{3}\right)=\sqrt{3}$

Now, substitute $x=\frac{6y}{9-y^2}$:

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

Divide the quadratic equation by $\sqrt{3}$:

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

Using the quadratic formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

$y=\frac{-2\sqrt{3}\pm \sqrt{12-4\left(1\right)(-9)}}{2}$
$y=\frac{-2\sqrt{3}\pm \sqrt{12+36}}{2}$
$y=\frac{-2\sqrt{3}\pm \sqrt{48}}{2}$
$y=\frac{-2\sqrt{3}\pm 4\sqrt{3}}{2}$
$y=-\sqrt{3}\pm 2\sqrt{3}$

This gives two values:
$y_1=\sqrt{3}$
$y_2=-3\sqrt{3}$

Since we assumed $0<y<3$:
- $y_1=\sqrt{3}\approx 1.732$ lies in $(0,3)$, so it is a valid solution.
- $y_2=-3\sqrt{3}\approx -5.196$ is outside $(0,3)$, so it is rejected.

Case 2: $-3<y<0$ (which means $x<0$)
Since $x<0$, ${cot}^{-1}\left(\frac{1}{x}\right)=\pi +{tan}^{-1}\left(x\right)$.
Substituting this into the equation gives:

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

Now, substitute $x=\frac{6y}{9-y^2}$:

$\frac{6y}{9-y^2}=-\frac{1}{\sqrt{3}}$
$6\sqrt{3}y=-9+y^2$
$y^2-6\sqrt{3}y-9=0$

Using the quadratic formula:

$y=\frac{6\sqrt{3}\pm \sqrt{108-4\left(1\right)(-9)}}{2}$
$y=\frac{6\sqrt{3}\pm \sqrt{108+36}}{2}$
$y=\frac{6\sqrt{3}\pm \sqrt{144}}{2}$
$y=\frac{6\sqrt{3}\pm 12}{2}$
$y=3\sqrt{3}\pm 6$

This gives two values:
$y_3=3\sqrt{3}+6\approx 11.196$ (Rejected, as it is not in $(-3,0)$)
$y_4=3\sqrt{3}-6\approx 5.196-6=-0.804$

Since $-3<3\sqrt{3}-6<0$, $y_4=3\sqrt{3}-6$ is a valid solution.

Step 3: Sum of all valid solutions
The valid solutions to the given equation for $0<\left|y\right|<3$ are:

$y=\sqrt{3}$ and $y=3\sqrt{3}-6$

Adding these solutions together:

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