Let $S=\{(x,y)\in \mathbb{C}^2:y^2=x\}$. Find the largest subset $U$ of $S$ such that for all $s\in U$, $s$ has a neighborhood $N_s$ which is homeomorphic to an open disk in $\mathbb{C}$.
If the parabola is real, then the whole parabola is homeomorphic to the real line by homeomorphism $\phi(y^2,y)=y$. So I consider the function $f:S\to\mathbb{C}$, where $f(r^2e^{i2\theta},re^{i\theta})=re^{i\theta}$. Clearly it is a bijection.
Claim: $f$ is continuous.
Let $\epsilon>0$, and $\delta<\epsilon$. Let $(r^2e^{i2\theta},re^{i\theta})$ be a point within $\delta$ distance of $(r_0^2e^{i2\theta_0},r_0e^{i\theta_0})$. Then,
$$|(r^2e^{i2\theta},re^{i\theta})-(r_0^2e^{i2\theta_0},r_0e^{i\theta_0})|^2<\delta^2$$ $$\implies|(r^2e^{i2\theta},re^{i\theta})-(r_0^2e^{i2\theta_0},r_0e^{i\theta_0})|^2<\epsilon^2$$ $$\implies|r^2e^{i2\theta}-r_0^2e^{i2\theta_0}|^2+|re^{i\theta}-r_0e^{i\theta_0}|^2<\epsilon^2$$ $$\implies|re^{i\theta}-r_0e^{i\theta_0}|^2<\epsilon^2$$ $$\implies|re^{i\theta}-r_0e^{i\theta_0}|<\epsilon$$
What's left is to check on which subset $f^{-1}$ is continuous. The professor gave a hint that the largest subset shouldn't be the entire set. But somehow I can't find a point where $f^{-1}$ is not continuous. Any suggestions or other approaches?
You are right, $S$ is homeomorphic to the complex plane.
Consider the projection $\pi_2 \colon \mathbb{C}^2 \to \mathbb{C},\; (w,z) \mapsto z$. This is clearly continuous, and your $f$ is the restriction of $\pi_2$ to $S$, hence also continuous. The inverse is easily given: $z \mapsto (z^2,z)$ is the inverse of $f$. Both component functions of $f^{-1}$ are continuous, hence $f^{-1}$ is continuous, and we see that $f$ is a homeomorphism. All maps considered here are even holomorphic, so $S$ is not only homeomorphic to the plane, it's biholomorphically equivalent to the plane.
Thus every point of $S$ has a neighbourhood homeomorphic to the open unit disk.