Let $p=\sum x$ and $q=\sum xy$. We have that $x^3-3x=y^3-3y=z^3-3z=a-3x-3y-3z$. So we get that $x,y,z$ are roots of polynomial $P(t)=t^3-3t+c$, where $c=3p-a$. From Vieta we get $p=0$ and $q=-3$ which means $c=-a$ and we should find possible values of $a=t^3-3t=x^3-3x$, where $x,y$ are roots of $x^2+xy+y^2=3$. Since $x\ne y\ne z=-x-y\ne x$, we get $x\ne y, x\ne -2y, y\ne -2x$. So, $x,y\ne \pm 1,\pm2$. Also since $y^2-y\cdot x+ (x^2-3)$ has roots, we get $|x|\le 2 \implies -2\le (x-1)^2(x+2)-2= x^3-3x=a=(x+1)^2(x-2)+2\le 2$. Since $x\ne \pm 1, \pm 2$, we get $a\in(-2,2)$ . Since we should also have $xy(x+y)\ne 0$ because of denominator can't be $0$. So, $a=0$ doesn't work. So, the answer is $a\in(-2,2)\setminus \{0\}$. @below thanks.

Note that $a \neq 0$ since sum of the two variables has to be $0$ which makes the denominator $0$

From the ratio identities we can easily obtain that $x^2+xy+y^2 = 3$, $y^2+yz+z^2 = 3$ and $z^2+zx+x^2 = 3$. Use any two of these three new equations to get $x + y + z = 0$. Plugging this back to the given equations, we get that $x$, $y$ and $z$ are the three distinct real roots of $m^3-3m+a = 0$. Also, note that since the denominators cannot be zero, $0$ cannot be a root of this equation and therefore $a \neq 0$. Now observe that for $a = 2$, $m^3-3m+2=(m-1)^2(m+2)$ and for $a = -2$, $m^3-3m-2=(m+1)^2(m-2)$. One may easily observe that using graph shifting about the $y$-axis, there cannot be three distinct real roots if $a > 2$ or $a < -2$. And finally, if $a \neq 2$ and $a \neq -2$, the graph of $m^3-3m+a$ will cut the $x$-axis at three points, so our final answer is $a \in (-2, 2) - \{0\}$.

With some algebratic calculations and equating the ratios, we get $x + y + z = 0$. If we write $y + z = -x$, then we have $x^3 +a = 3x$. At the end, we have a equation of the third degree $x^3 -3x +a =0$. The situations are symmetric. So : $y^3 -3y +a =0$. $z^3 -3z + a=0$. Notice that $ a\neq{0}$. Due to $x, y, z$ are different real numbers, the equation $t^3 -3t +a=0$ must have three distinct real roots. And these roots are $x, y$ and $z$. If $ a \in (-2,2)-$$ \{ 0 \} $, then, the equation $t^3 -3t +a $ has three distinct real roots. If $ a= \{2,2\} $, then the equation has two distinct real roots. If $ a<-2 $, then the equation has just one real root. The case of $ a>2 $ is same with $ a<-2 $. So, the answer is : $a \in (-2,2)- \{0\}$. $\square$