Apparently the expression of the result changes depending on the sign of $\omega$, I'm not sure why.
$$\mathcal{F}[f](\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{xe^{-i\omega x}}{x^2+x+1}dx$$
So I consider this complex function:
$$\frac{ze^{-i\omega z}}{z^2+z+1}$$
It has poles at $-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$. I integrate over a half circle of radius $R$ in the upper half of the complex plane, closed by a line through the real numbers from $-R$ to $R$. Since:
$$\frac{|Re^{-i\theta}e^{-i\omega Re^{i\theta}}|}{|R^2e^{i2\theta} + Re^{i\theta} + 1|} \leq \frac{Re^{\omega R \sin(\theta)}}{R^2 - R - 1}$$
Then if C is the circular part of the path: $$|\int_{C}f(z)dz| \leq \int_{C}|f(z)|dz \leq \int_{0}^{\pi} \frac{Re^{\omega R \sin(\theta)}}{R^2 - R - 1} d\theta$$
$$ |\int_{C}f(z)dz| \leq \frac{R}{R^2 - R - 1} \int_{0}^{\pi} e^{\omega R \sin(\theta)} d\theta \rightarrow 0 \space \space (R \rightarrow \infty) $$
So:
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{xe^{-i\omega x}}{x^2+x+1}dx = \frac{2\pi i}{\sqrt{2\pi}} Res(f(z), -\frac{1}{2} + i \frac{\sqrt{3}}{2}) $$
Where I used the residue over the pole at $\frac{1}{2} + i \frac{\sqrt{3}}{2}$ because I chose the circle in the upper plane, and that's it. Where does the sign of $\omega$ change the result?
Thanks.
Note that for your range of integration, $e^{\omega R \sin \theta}$ only tends to zero provided $\Re(\omega) < 0$, which is implicitly the bound you're using.
To solve this, for $\Re(\omega) > 0$ you instead close your contour in the lower half plane, and this way you pick up the other residue.