I haven't seen a proof of this here on Math Stack Exchange, and searching the internet, I couldn't find it clearly stated somewhere. So I'll state the claim and provide a sketch for a proof. If you find that there's some invalid step in the proof, please point it out. Thanks.
Claim:
Let $a \in \Bbb C$ and $D$ be a domain in $\Bbb C$ such that $ a\in D$. Let $f,g: D \to \Bbb C$ be analytic non-identically-zero functions such that $f(a) = g(a) = 0$. On the condition that $\lim_{z \to a }\frac{f'(z)}{g'(z)}$ exists, we have that $\lim_{z \to a}\frac{f(z)}{g(z)}$ exists and:
$$L:=\lim_{z \to a}\frac{f(z)}{g(z)} = \lim_{z \to a }\frac{f'(z)}{g'(z)}=:L'$$
As $L'$ exists, we have $\lim_{z \to a}\frac{|f'(z)|}{|g'(z)|} =: M < \infty$, hence there exists some $\delta > 0$ such that $|f'(z)| \le (1+M)|g'(z)|$ in $B(a,\delta)$. If $f'$ is identically zero then there's nothing to prove. We also can't have $g'$ identically zero. This and the inequality imply that the order of $a$ as a zero of $f'$ is at least the order of $a$ as a zero of $g'$. Therefore the order $m$ of $a$ as a zero of $f$ is at least $n$, the order of $a$ as a zero of $g$. If $m=n$, we get through Taylor expansion that $L = f^{(m)}(a)/g^{(m)}(a)$. Otherwise $L=0$. One can also check that this is what we obtain for $L'$.
$\square$.