The answer:
$$\frac{2(-1)^n \cdot n!}{\sqrt{3}r^{n+1}}\sin(n+1)\theta,$$
where $r=\sqrt{x^2+x+1}$ and $\theta=\cot^{-1}\frac{2x+1}{\sqrt{3}}$.
How I have tried:
$$y=\frac{1}{x^2+x+1+\frac{1}{4}-\frac{1}{4}}$$
$$y=\frac{1}{(x^2+x+\frac{1}{4}-\frac{3}{4})}$$
$$y=\frac{1}{(x+\frac{1}{2})^2-\frac{3}{4}}$$
$$y=\frac{4}{(2x+1)^2-3}$$
I don't know how to proceed further, someone please help.
HINT:
Let $x^2+x+1=(x-a)(x-b)$
So, we can write $\displaystyle a,b=\frac{-1\pm\sqrt3i}2$
$$\implies\frac1{x^2+x+1}=\frac1{a-b}\left(\frac1{x-b}-\frac1{x-a}\right)$$
Now from this and many others, $$\frac{d^n(1/(x-a))}{dx}=(-1)^n\frac{n!}{(x-a)^{n+1}}$$
For $\displaystyle x-a=x-\left(\dfrac12+i\dfrac{\sqrt3}2\right)$ set $x-\dfrac12=r\cos\phi,$ and $-\dfrac{\sqrt3}2=r\sin\phi$
and apply de Movire Theorem