Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$. Let $z \in D \setminus \set {x \in D: \map g z = 0}$. By the Alternative Differentiability Condition, it follows that there exists $r_0 \in \R_{>0}$ such that for all $h \in \map {B_{r_0} } 0 \setminus \set 0$: :$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilon_f} h}$ :$\map g {z + h} = \map g z + h \paren {\map {g'} z + \map {\epsilon_g} h}$ where $\epsilon_f, \epsilon_g: \map {B_{r_0} } 0 \setminus \set 0 \to \C$ are continuous functions that converge to $0$ as $h$ tends to $0$. From Complex-Differentiable Function is Continuous, it follows that $g$ is continuous at $z$. Then there exists $r_1 \in \R_{>0}$ such that for all $h \in \map {B_{r_1} } 0$, we have $\size {\map g {z + h} - \map g z} < \size {\map g z}$. Then by Backwards Form of Triangle Inequality: :$\map g {z + h} \ne 0$ for all $h \in \map {B_{r_1} } 0$ Put $r = \min \set {r_0, r_1}$. Then for all $h \in B_r \setminus \set 0$: {{begin-eqn}} {{eqn | l = \dfrac {\map f {z + h} } {\map g {z + h} }       | r = \dfrac {\map f z} {\map g z} + \dfrac {\map f {z + h} - \map f z} {\map g {z + h} } - \map f z \dfrac {\map g {z + h} - \map g z} {\map g z \map g {z + h} }       | c = adding and subtracting the same quantity }} {{eqn | r = \dfrac {\map f z} {\map g z} + \dfrac {h \paren {\map {f'} z + \map {\epsilon_f} h} } {\map g z + h \paren {\map {g'} z + \map {\epsilon_g} h} } - \dfrac {\map f z h \paren {\map {g'} z \map {\epsilon_g} h} } {\map g z \paren {\map g z + h \paren {\map {g'} z + \map {\epsilon_g} h} } } }} {{eqn | r = \dfrac {\map f z} {\map g z} + h \dfrac {\map {f'} z \map g z + \map g z \map {\epsilon_f} h - \map f z \map {g'} z - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}       | c = with $\map {\epsilon_0} h = h \map {g'} z \paren {\map {g'} z + \map {\epsilon_g} h}$ }} {{eqn | r = \dfrac {\map f z} {\map g z} + h \paren {\dfrac {\map {f'} z \map g z - \map f z \map {g'} z} { {\paren {\map g z}^2 } } + \map \epsilon h} }} {{end-eqn}} where $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ is defined by: :$\map \epsilon h = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \dfrac {\map g z \map {\epsilon_f} h - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}$ From Combination Theorem for Continuous Functions, it follows that $\epsilon$ is continuous. From Combination Theorem for Limits of Functions, it follows that: {{begin-eqn}} {{eqn | l = \lim_{h \mathop \to 0} \map \epsilon h        | r = \lim_{h \mathop \to 0} \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \lim_{h \mathop \to 0} \dfrac {\map g z \map {\epsilon_f} h - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h} }} {{eqn | r = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \map g z \lim_{h \mathop \to 0} \dfrac {\map {\epsilon_f} h} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \map f z \lim_{h \mathop \to 0} \dfrac {\map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h} }} {{eqn | r = 0 }} {{end-eqn}} Then the Alternative Differentiability Condition shows that: :$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$ {{qed}}