Let $D\subseteq\mathbb C$ be open, $c\in D$ and $f,g:D\to\mathbb C$ holomorphic. Assume that $g(z)$ has a zero in $c$ of order 2. Show that $$\text{Res}_c\left(\frac{f}{g}\right)=\frac{6f'(c)g''(c)-2f(c)g'''(c)}{3(g''(c))^2}$$
How do I start here? At first I tried to look at the different cases where $c$ is also a zero of $f$ of order $n=2$ (or $n=1$) and then take $$\text{Res}_c=\frac{d}{dz}\mid_{z=c}\left (z-c)^n\cdot\frac{f(z)}{g(z)}\right)$$ but that approach doesn't look promising at all.
Write $$f(c+w)=a_0+a_1w+\cdots$$ and $$g(c+w)=b_2w^2+b_3w^3+\cdots.$$ Then $$\frac{f}g(c+w)=w^{-2}\frac{a_0+a_1w+\cdots}{b_2(1+(b_3/b_2)w+\cdots)} =\frac{a_0+(a_1-a_0b_3/b_2)w+\cdots}{b_2w^2}$$ so the residue is $$\frac{a_1b_2-a_0b_3}{b_2^2}.$$ Now use $a_0=f(c)$, $a_1=f'(c)$, $2b_2=g''(c)$, $6b_3=g'''(c)$.