If one of them , say $ x=0$, then we have at least one of the others, say $ y=0$ also, so $ (0,0,\frac{1}{\sqrt{3}})$ and its premutations are solutions. Now for $ x,y,z \neq 0$, we have $ \frac{x^2+y^2+z^2}{3} \ge \frac{(|x|+|y|+|z|)^2}{3^2}$ $ \Leftrightarrow |x|+|y|+|z| \le 1$ So now we have $ x^2y^2+y^2z^2+z^2x^2=xyz(x+y+z)^3 \le |x||y||z|(|x|+|y|+|z|)$ But, $ \sum (x^2y^2z^0) \ge \sum (|x|^2|y||z|)$ by muirhead. So equality holds $ \Rightarrow x=y=z$ is positive , which gives $ x=y=z=\frac{1}{3}$ as solution after checking. So $ (x,y,z)=(\frac{1}{3},\frac{1}{3},\frac{1}{3}), (0,0,\frac{1}{\sqrt{3}}),(0,\frac{1}{\sqrt{3}},0),(\frac{1}{\sqrt{3}},0,0)$ are the only solutions.

What about $ \Big( \frac{-1}{3}, \frac{-1}{3}, \frac{-1}{3} \Big)$ and $ \Big(0, 0, \frac{-1}{\sqrt 3}\Big)$?

djuro wrote: What about $ \Big( \frac { - 1}{3}, \frac { - 1}{3}, \frac { - 1}{3} \Big)$ and $ \Big(0, 0, \frac { - 1}{\sqrt 3}\Big)$? Sorry , correction to my solution: The case $ x=0$ gives $ y=0, z^2=\frac{1}{3}$, so $ (0, 0, \frac {-1}{\sqrt 3})$ and its premutations are solutions also. On the second line from the bottom, it should be equality holds $ \Rightarrow |x|+|y|+|z|=|x+y+z| \Rightarrow x,y,z$ are all positive or all negative, and $ |x|=|y|=|z|$, these gives $ x=y=z=\frac{1}{3}$ or $ x=y=z=-\frac{1}{3}$ after checking. So $ (x,y,z)=(\frac{1}{3},\frac{1}{3},\frac{1}{3}), (-\frac{1}{3},-\frac{1}{3},-\frac{1}{3}),(0, 0, \frac {1}{\sqrt 3})(0, 0, \frac {-1}{\sqrt 3})$ and their premutations are the solutions.

This is Indonesia IMO TST 2004.. $ (x+y+z)^2 \le 3(x^2+y^2+z^2)=1$ $ xyz(x+y+z)\ge xyz(x+y+z)^3=x^2y^2+y^2z^2+z^2x^2 \ge xyz(x+y+z)$ I think this is the shortest path..

Another solution $ xyz(x+y+z)^{3}=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}=3(x^{2}+y^{2}+z^{2})(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}) \ge 3xyz(x+y+z)(x^{2}+y^{2}+z^{2})$

It's an old problem of Poland 1997

$3(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)=xyz(x+y+z)^3)$. In the left hand we have only positive numbers, so we will work with positive $x,y,z$(only have to take $|x|,|y|,z$and permutations. The main ideea( i won't study particular cases) is $3(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)\ge xyz(x+y+z)^3)$. The inequality is not hard, after expending we get:$3 \displaystyle\sum\limits_{\text{cyc}}(x^4y^2+x^2y^4)+3x^2y^2z^2\ge3 \displaystyle\sum\limits_{\text{cyc}}(x^3y^2z+x^3yz^2)+x^4yz+xy^4z+xyz^4$. We can aply AM-GM for the numbers $x^4y^2+x^2y^2z^2\gex^3y^2z$. The rest of the inequality is easy to prove using Muirhead. ($4\ge3, 4+2\ge3+2, 4+2+0=3+2+1$, and $4\ge4, 4+2\ge4+1, 4+2+4+1+1$,etc)

We claim the solutions are $(0,0,\pm1/\sqrt3)$, $\pm1(1/3,1/3,1/3)$, and permutations. These are easily seen to work. Now, using AM-GM and the triangle inequality, \begin{align*} |xyz(x+y+z)^3|&=|x^2y^2+y^2z^2+z^2x^2|\\ |xyz||(x+y+z)^3|&=x^2y^2+y^2z^2+z^2x^2\\ |xyz||x+y+z|^3&\ge |xyz|(|x|+|y|+|z|)\\ |xyz||x+y+z|^3&\ge|xyz||x+y+z|\\ |xyz||x+y+z|((x+y+z)^2-1)&\ge0. \end{align*}Now, if $xyz(x+y+z)=0$, then it follows two of $x,y,z$ are $0$, from which we get the solutions $(0,0,\pm1/\sqrt3)$. Otherwise, $(x+y+z)^2\ge1$. But then, $xy+yz+zx\ge\frac13=x^2+y^2+z^2\ge xy+yz+zx$, so from the equality case of the trivial inequality, $x=y=z$. This gives $\pm1(1/3,1/3,1/3)$. $\blacksquare$