Finding a representation of an element of Fibonacci sequence using power series. I need help finding the power series of $f\left(x\right)=\frac{-1}{x^{2}+x-1}$ probably for $x=0$.
I got a sequence defined as $a_{0}=1,\ a_{1}=1\ \ a_{n}=a_{n-1}+a_{n-2}$ for $n\ge 2$. I found out that the radius of convergence of the series $f\left(x\right)=\sum_{n=0}^{\infty}a_{n}x^{n}$ is bigger than $\frac{1}{2}$. Afterwards I found for all $|x|<\frac{1}{2}$ that $$f\left(x\right)=\frac{-1}{x^{2}+x-1}.$$ And now I think I have to find out that the power series of $\frac{-1}{x^{2}+x-1}$ is $$\sum_{n=0}^{\infty}\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}}{\sqrt{5}}x^{n}$$ and then I can conclude from the similarities of power series that $a_{n}=\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}}{\sqrt{5}}$.
I started by getting to this equality: $$f\left(x\right)=\frac{1}{\left(x-\frac{-1-\sqrt{5}}{2}\right)\left(x-\frac{-1+\sqrt{5}}{2}\right)}\\=\frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{-1+\sqrt{5}}{2}}-\frac{1}{x-\frac{-1-\sqrt{5}}{2}}\right).$$ I wanted to show out that $$\frac{1}{x-\frac{-1+\sqrt{5}}{2}}=\sum_{n=0}^{\infty}\left(\frac{1+\sqrt{5}}{2}\right)^{n}x^{n}, \\ \frac{1}{x-\frac{-1-\sqrt{5}}{2}}=\sum_{n=0}^{\infty}\left(\frac{1-\sqrt{5}}{2}\right)^{n}x^{n},$$ but I can't manage to do that, I tried expanding the Taylor series around $0$ but it's not working for me.
Any hints or help would be great. Thank you!

$$\frac{1}{x-\frac{-1+\sqrt{5}}{2}}=\frac{2}{1-\sqrt{5}}\frac{1}{1-x\cdot\frac{2}{\sqrt{5}-1}}$$
Now let’s use rationalising/conjugacy technique: $$\frac{2}{\sqrt{5}-1}=\frac{2(\sqrt{5}+1)}{(\sqrt{5})^2-1^2}=\frac{\sqrt{5}+1}{2}$$
So you get: $$-\frac{1+\sqrt{5}}{2}\cdot\frac{1}{1-x\cdot\frac{1+\sqrt{5}}{2}}$$
Can you proceed with a geometric series?
Two notes: You’ve dropped a minus sign in your factorisation of $f$, and you’ve dropped a power - those two series you ask about should feature $\varphi^{n\color{red}{+1}}x^n$.
