Find a smallest number $\alpha$, such that for all $x,y,z$ (not all of which are positive) inequality $$\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$$
Let $f(t)=t^2-t+1$. Then $f(t) \ge \frac34$.
If $x=0, y=z=\frac12$, then $$\alpha\ge \frac{16}9$$
On
From your $a(x^2-x+1)^3\ge x^6-x^3+1 $, since $(x+1)(x^2-x+1) =x^3+1 $ and $(x^3+1)(x^6-x^3+1) =x^8+1 $,
$a(\frac{x^3+1}{x+1})^3\ge \frac{x^9+1}{x^3+1} $ or $a\ge \frac{(x^9+1)(x+1)^3}{(x^3+1)^4} $.
According to Wolfy, this has a maximum of $2.1547$ at $x \approx 0.435421 $ and $2.29663 $. These are the roots of $0 = x^6-2 x^5-x^4+x^2+2 x-1 $.
The exact roots are $-1, 1, \frac12 (1\pm\sqrt{2} 3^{1/4}+\sqrt{3}), \frac12 (1\pm i\sqrt{2} 3^{1/4}-\sqrt{3}) $.
So $a$ must be at least $2.1547$.
In the starting formulation the answer is $\frac{16}{9}$.
If $x$, $y$ and $z$ are non-positives so after replacing $x\rightarrow-x$... we need to prove that $$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2+z+1)\geq x^2y^2z^2-xyz+1$$
which is true for all non-negatives $x$, $y$ and $z$ because we'll prove now that even $$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$.
If $x\leq0$, $y\leq0$ and $z\geq0$ we need to prove that $$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2-z+1)\geq x^2y^2z^2-xyz+1$$ for non-negatives $x$, $y$ and $z$, which follows from $$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$ again.
Now we'll prove it:
$$\begin{align} &16(x^2+x+1)(y^2-y+1)(z^2-z+1)-9(x^2y^2z^2+xyz+1) \\&\phantom{aaa}=(16(y^2-y+1)(z^2-z+1)-9y^2z^2)x^2+(16(y^2-y+1)(z^2-z+1)-9yz)x \\ &\phantom{aaaaa}+16(y^2-y+1)(z^2-z+1)-9 \\&\phantom{aaa} \geq\left(16\left(\frac{3}{4}y^2+\left(\frac{y}{2}-1\right)^2\right)\left(\frac{3}{4}z^2+\left(\frac{z}{2}-1\right)^2\right)-9y^2z^2\right)x^2+ +(3y\cdot3z-9yz)x \\&\phantom{aaaaa}+ 16\left(\frac{3}{4}+\left(\frac{1}{2}-y\right)^2\right)\left(\frac{3}{4}+\left(\frac{1}{2}-z\right)^2\right)-9 \\&\phantom{aaa}\geq0 \end{align}$$