First, the give condition $a^2 + ab + b^2 = 1$ can be rewritten as \begin{align} (a+b)^2 - ab &=1 &\implies &ab = (a+b)^2 -1 \\ (a-b)^2 + 3ab &=1 &\implies &ab = \frac{1}{3}-\frac{1}{3}(a-b)^2 \end{align} Similarly, rewriting the target as \begin{align} a^2 - ab + b^2 &= (a+b)^2 - 3ab &=(a+b)^2 - 3((a+b)^2-1)&=-2(a+b)^2 +3&\le 3 \\ a^2 - ab + b^2 &= (a-b)^2 + ab &=(a-b)^2 + \Big(\frac{1}{3}-\frac{1}{3}(a-b)^2\Big)&=\frac{2}{3}(a-b)^2+\frac{1}{3}&\ge\frac{1}{3} \end{align} Hence, the answer is $\boxed{\Big[\frac{1}{3}, 3\Big]}$. When $a=-b$, i.e. $(a, b)=(\pm 1, \mp1)$, the target reaches maximum value. When $a=b=\frac{\sqrt{3}}{3}$, the target reaches minimal value.