So suppose we have $X=\{(x,y)\in \mathbb{C} : y^2=x^2+x^3\}$ and i am asked to see that this is not a Riemann surface. Well we know that it is not going to be a smoth affine plane curve because at the point $(0,0)$ both the derivatives are $0$ but this is not enough, how can one ensure that there is possibly no way we cant find complex chart for this? I have tried supposing we can find a chart near $(0,0)$, so that we have $\phi^{-1}(z)=(\phi_1(z),\phi_2(z))$ near that point such that $\phi_2^2(z)=\phi_1^3(z)+\phi_1^2(z)$ but i cant seem to find a contradiction.Maybe seing something with the function $F$ that defines this locus set , since $\frac{\partial F}{\partial y} = 2\phi_2(z)\phi_2'(z)=0$ and $\frac{\partial F}{\partial x}=3\phi_1(z)^2\phi_1'(z)+2\phi_1(z)\phi_1'(z)=0$ so we get additional information that $3\phi_1(z)^2+2\phi_1(z)=0$, so that $\phi_1(z)=-\frac{2}{3}$ and this is a contadiction.Would this for example work ?
Thanks in advance.
There are two pieces of smooth curve (= two Riemann surfaces) passing through $(0,0)$
$$y^2=x^2(1+x), \qquad y = \pm x \sqrt{1+x}$$
There is no "chart" from a neighborhood of $0$ to two copies of a neighborhood of $0$.
The function $$y+x\sqrt{1+x}$$ vanishes on one but not on the other which means that the ring of functions from the curve to $\Bbb{C}$ meromorphic near $(0,0)$ is not an integral domain (it is the product of two fields, two copies of the field of meromorphic functions near $0\in \Bbb{C}$).