Let $\text{x},\text{y},\text{z}$  be non zero real numbers that satisfy the system of equations:$(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)=xyz$, and  $(x^4+x^2{y}^2+y^4)(y^4+y^2{z}^2+z^4)(z^4+z^2{x}^2+x^4)=x^3{y}^3{z}^3$  
 

Since  $xyz\ne 0,$ we can divide the second relation by the first. Observe that 
 $x^4+x^2{y}^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2).$
Holds for any x,y .Thus we get  $(x^2-xy+y^2)(y^2-yz+z^2)(z^2-zx+x^2)=x^2{y}^2{z}^2$
However , for any number x,y, we have $x^2-xy+y^2>\left|xy\right|$
Since     $x^2{y}^2{z}^2=\left|xy\right|\left|yz\right|\left|zx\right|,weget$$\left|xy\right|\left|yz\right|\left|zx\right|=(x^2-xy+y^2)(y^2-yz+z^2)(z^2-zx+x^2)\ge 0\left|xy\right|\left|yz\right|\left|zx\right|$
This is possible only if
$x^2-xy+y^2=\left|xy\right|,y^2-yz+z^2=\left|yz\right|,z^2-zx+x^2=\left|zx\right|$ ,
Hold simultaneously, However $\left|xy\right|=\pm xy.$ If $x^2-xy+y^2=-xy,$  then $x^2+y^2=0$  giving   $x=y=0.$
Since we are looking for nonzero  $x,y,z$ we conclude that  $x^2-xy+y^2=xy$ which is  same as $x=y.$ .Using the other relations, we also get  $y=z$ and $z=x$. The first equation  now gives  $27x^6=x^3$.This gives  $x^3=\frac{1}{27}(\text{since\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x\ne 0),orx=\frac{1}{3}$.We thus have   $x=y=z=\frac{1}{3}.$ These also satisfy the second relation, as may be verified