If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$? 
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$. 

So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer and longer. Am I on the right track? Thanks!

Let $\omega$ be a complex cube root of $1$. Then Partial Fraction representation of $f(x)$ is given by
$f(x) = \dfrac{1}{x^2+x+1} = \dfrac{1}{(x-\omega)(x-\omega^2)} = \dfrac{1}{\omega - \omega^2}\Big(\dfrac{1}{x-\omega} - \dfrac{1}{x - \omega^2}\Big)$.
Find successive derivatives to show that
$f^{(36)}(x) = \dfrac{1}{\omega - \omega^2}(36! (x-\omega)^{-37} - 36! (x - \omega^2)^{-37})$.
Let $x = 0$ and use $\omega^3 = 1$.
