Let $f,g \in H(D(a;r))$ and assume that $f$ has a zero of order $m$ and $g$ a zero of order $m+1$ at $a$. Prove that $$\mathrm{Res}\left(\frac{f(z)}{g(z)}, a\right)=(m+1)\frac{f^{(m)}(a)}{g^{(m+1)}(a)}$$
What I did first was writing
$$f(z)=\frac{h_1(z)}{(z-a)^m} \; g(z)=\frac{h_2(z)}{(z-a)^m}$$ where both $h_1, h_2$ are holomorphic on the open disc and $h_1(a)\neq 0, h_2(a)\neq 0$. Then, by the Cauchy Formula for derivatives
$$\begin{align*} (m+1)\frac{f^{(m)}(a)}{g^{(m+1)}(a)} &= (m+1) \frac{\frac{m!}{2\pi i}\int_{\gamma(a;r/2)}\frac{h_1(z)}{(z-a)^{m+1}} dz }{\frac{(m+1)!}{2\pi i}\int_{\gamma(a;r/2)}\frac{f(z)}{(z-a)^{m+2}} dz}\\ &= \frac{(m+1)!}{(m+1)!}\int_{\gamma(a;r/2)}\frac{f(z)}{g(z)} dz\\ &= \int_{\gamma(a;r/2)} \frac{f(z)}{g(z)} dz\\ &= 2\pi i\cdot \mathrm{Res} \left( \frac{f(z)}{g(z)} ; a\right) \end{align*}$$ so I get and extra factor $2\pi i$ and I don't know what I did wrong. Any help would be great with that.
I can't follow what you're doing. Me, I would let $f(z) = (z-a)^m(c_m + c_{m+1}(z-a) +\cdots),$ $ g(z) = (z-a)^{m+1}(d_{m+1}+ d_{m+2}(z-a)+ \cdots),$ where $c_m, d_{m+1} \ne 0.$ Thus for $z$ near $a,z\ne a,$
$$\frac{f(z)}{g(z)} = \frac{1}{z-a}\cdot \frac{c_m + c_{m+1}(z-a) +\cdots}{d_{m+1} +d_{m+2}(z-a)+ \cdots} = \frac{1}{z-a}\cdot h(z).$$
Now $h$ is the quotient of holomorhic functions that are both nonzero at $a,$ so $h(z)$ extends to be holomorphic near $a,$ with $h(z) = h(a) + h'(a)(z-a) + \cdots$ there. So
$$\frac{f(z)}{g(z)} = \frac{h(a)}{z-a} + h'(a) + \cdots.$$
Thus the residue of $f/g$ at $a$ is $h(a) = c_{m}/d_{m+1}.$ Since $c_m,d_{m+1}$ are Taylor coefficients of $f,g,$ we get the result.