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!
We can write:

$$1+ x + x^2 = \frac{1-x^3}{1-x}$$

Therefore:

$$f(x) = \frac{1-x}{1-x^3} $$

We can then expand this in powers of $x$:

$$f(x) = (1-x)\sum_{k=0}^{\infty}x^{3 k}$$

which is valid for $\left|x\right|&lt;1$. The coefficient of $x^{36}$ is thus equal to $1$, so the 36th derivative at $x = 0$ is $36!$ .
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 $$a_{36} = b_{36}\cdot 36! = 36!$$