Let $G\subset \mathbb{C}$ be a domain and $f:G\to \mathbb{C}$ be continuously differentiable. Let $H\subset G$ dense s.t. $f$ is holomorphic in every $z\in H$. Then $f$ is already holomorphic on $G$.
The hint is: I have to show that the set $A:=\{z\in G | f \text{ holomorphic in } z\}$ is open and closed in $G$.
I have no idea how to prove this.
Let $z\in H$
"A open": I don't think I can use the fact that $f$ must be analytical in a ball around $z$ since I need holomorphy on a neighbourhood a priori. So, I have no idea here.
"A closed": Are the partial derivatives of a continuously differentiable function continuous as well? Then the Cauchy-Riemann equations should hold due to density of $H$ in $G$.
Let me make a suggestion how to understand the question: We assume that $f$ (considered as a function from an open subset $G$ of $\mathbb{R}^2$ to $\mathbb{R}^2$) is continuously differentiable on $G$ and complex differentiable in all points of $H$.
This means that $f = (u,v) = u + iv$ with functions $u,v : G \to \mathbb{R}$ such that
(1) $u,v$ have continuous partial (real) derivatives on $G$
(2) These partial derivatives satisfy the Cauchy Riemann equations in all $(x,y) = x +iy \in H$.
Then, as you stated correctly, the Cauchy Riemann equations must be satisfied in all points of $\overline{H} = G$, where $\overline{\phantom{H}}$ denotes the closure in $G$. This is due to continuity of the partial derivatives: Each point of $G$ is the limit of a sequence in $H$.
More generally, if you have a continuously differentiable function $f : G \to \mathbb{C}$, then the above argument shows that the set of $z \in G$ in which $f$ is complex differentiable is closed in $G$. In general this set is not open. Consider for example $f : \mathbb{C} \to \mathbb{C}, f(x,y) = x^2 + y^2$ (i.e. $u(x,y) = x^2 + y^2, v(x,y) = 0)$. Then the Cauchy Riemann equations are satisfied exactly in $(0,0)$.