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!

A non power/taylor series method using Leibniz' generalize product rule:
$$ (fg)^{n}(x) = \sum_{r = 0}^{n} \binom{n}{r} f^{n-r}(x) g^{r}(x)$$
Let $f(x) = \dfrac{1}{x^2 + x + 1}$ and $g(x) = x^2 + x +1 $
Then $f(x) g(x) = 1$. We apply the rule to $fg$.
If $a_k = f^{k}(0)$ the we get the recurrence relation
$$a_k + ka_{k-1} + k(k-1)a_{k-2} = 0$$
With $a_{0} = 1$ and $a_1 = -1$
Now let $b_k = \dfrac{a_k}{k!}$, then we get
$$b_k + b_{k-1} + b_{k-2} = 0$$
with $b_{0} = 1$ and $b_{1} = -1$.
Thus we get that the $b_i$ are
$$1, -1, 0, 1, -1, 0, 1, \dots$$
We see that $b_{3k} = 1$, $b_{3k+1} = -1$ and $b_{3k+2} = 0$.
Thus $$a_{36} = b_{36}\cdot 36! = 36!$$
