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

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 $\left|x\right|<1$. The coefficient of $x^{36}$ is thus equal to $1$, so the 36th derivative at $x = 0$ is $36!$ .

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 $\omega^3 = 1$.

Use $$x^2+x+1=\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)$$

By this hint we obtain: $$\left(\frac{1}{x^2+x+1}\right)^{(36)}_{x=0}=\left(\frac{1}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)}\right)^{(36)}_{x=0}=$$ $$=\left(\frac{1}{\sqrt3i}\left(\frac{1}{x+\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{1}{x+\frac{1}{2}+\frac{\sqrt3}{2}i}\right)\right)^{(36)}_{x=0}=$$ $$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)_{x=0}=$$ $$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)=\frac{1}{\sqrt3i}\left(\frac{36!}{\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{36!}{\frac{1}{2}+\frac{\sqrt3}{2}i}\right)=36!$$