Let's fight inequality

A trivial solution to the system of equations is $x=y=z=0$ or $(x,y,z)=(0,0,0)$ . For $x, y, z \ne 0$ , we can consider the reciprocal of each of equation as follows.

$\begin{cases} \dfrac{6x^{2}}{1+9x^{2}}=y & \implies \dfrac 1{6x^2} + \dfrac 32 = \dfrac 1y & ...(1) \\ \dfrac{6y^{2}}{1+9y^{2}}=z & \implies \dfrac 1{6y^2} + \dfrac 32 = \dfrac 1z & ...(2) \\ \dfrac{6z^{2}}{1+9z^{2}}=x & \implies \dfrac 1{6z^2} + \dfrac 32 = \dfrac 1x & ...(3) \end{cases}$

$(1)+(2)+(3)$ and rearrange:

$\begin{aligned} \frac 1{6x^2} - \frac 1x + \frac 32 + \frac 1{6y^2} - \frac 1y + \frac 32 + \frac 1{6z^2} - \frac 1z + \frac 32 & = 0 \\ \frac 1{x^2} - \frac 6x + 9 + \frac 1{y^2} - \frac 6y + 9 + \frac 1{z^2} - \frac 6z + 9 & = 0 \\ \left(\frac 1x - 3\right)^2 + \left(\frac 1y - 3\right)^2 + \left(\frac 1z - 3\right)^2 & = 0 \end{aligned}$

Since $\left(\dfrac 1x - 3\right)^2 \ge 0$ , $\left(\frac 1y - 3\right)^2 \ge 0$ and $\left(\frac 1z - 3\right)^2 \ge 0$ , for the $LHS=RHS=0$ , all the three square factors are simultaneously equal to 0. That is $x=y=z = \frac 13$ or $(x,y,z)=\left(\frac 13,\frac 13,\frac 13 \right)$ .

Therefore, the sum of all $x+y+z$ is $0+0+0+\frac 13 + \frac 13 + \frac 13 = \boxed{1}$ .

It's obvious that $x, y, z$ are non-negative numbers smaller than 1. Furthermore, since $(3x-1)^{2}\geq 0$ , we have

$9x^{2}+1\geq 6x$

$1\geq \frac{6x}{9x^{2}+1}$

$x\geq \frac{6x^{2}}{9x^{2}+1}$

Thus, $x\geq y\geq z\geq x$

$\Rightarrow x=y=z$

$\Rightarrow x=y=z=\frac{1}{3}$ , or $x=y=z=0$

Therefore the answer is $0+3(\frac{1}{3})=1$