power series for $\frac1{x^2+x+1}$ So I am trying to find the sequence $(a_n)_{n\geq0}$ such that 
$$\frac1{x^2+x+1}=\sum_{n\geq0}a_nx^n$$
My attempts:
I defined $$A(x)=\frac1{x^2+x+1}$$
Hence $$A(x)=\sum_{n\geq0}a_nx^n$$
And so $$(x^2+x+1)\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_nx^{n+2}+\sum_{n\geq0}a_nx^{n+1}+\sum_{n\geq0}a_nx^n=1$$
$$\sum_{n\geq0}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$a_0(x^2+x+1)+\sum_{n\geq1}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$x=0$:
$$a_0=1$$
So
$$x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$\frac{x^2+x+1+a_1(x^3+x^2+x)+\sum_{n\geq2}a_n(x^{n+2}+x^{n+1}+x^n)}x=\frac1x$$
$$x+1+a_1(x^2+x+1)+\sum_{n\geq2}a_n(x^{n+1}+x^{n}+x^{n-1})=0$$
$x=0$:
$$a_1=-1$$
One more time:
$$x^2+x+1-x^3-x^2-x+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=1$$
$$-x^3+a_2(x^4+x^3+x^2)+\sum_{n\geq3}a_n(x^{n+2}+x^{n+1}+x^n)=0$$
Divide both sides by $x^2$:
$$-x+a_2(x^2+x+1)+\sum_{n\geq3}a_n(x^{n}+x^{n-1}+x^{n-2})=0$$
$x=0$:
$$a_2=0$$
How do I find $a_n$? Am I doing things right so far? Thanks.

You could have continue your way.
Starting from
$$\sum_{n=0}^\infty a_nx^{n+2}+\sum_{n=0}^\infty a_nx^{n+1}+\sum_{n=0}^\infty a_nx^n=1$$ you already found $a_0=1$ and $a_1=-1$.
Now, consider a given degree $m >1$. You then have the recurrence relation
$$a_{m-2}+a_{m-1}+a_m=0$$ and the roots of the characteristic equation 
$$r^2+r+1=0 \implies r_\pm=\frac{-1 \pm i \sqrt 3} 2$$ making by the end (I let to you the intermediate steps)
$$a_m=\cos \left(\frac{2 \pi  m}{3}\right)-\frac{1}{\sqrt{3}}\sin \left(\frac{2 \pi 
   m}{3}\right)$$ making $a_{3m}=1$, $a_{3m-1}=0$, $a_{3m+1}=-1$.
