$f$ is a polynomial function; $a$, $b$ are complex variables.
$$X =\{(a,b)|f(a,b)=0\}\subset \mathbb C\times\mathbb C$$
When $f=0$, the complex gradient of $f$ is non zero:
$$\forall (a,b), s.t. (\frac{\partial f}{\partial a}\frac{\partial f}{\partial b})(a,b)\neq 0 \ \ \text{when} f(a,b)=0.$$
Prove that $X$ is a Riemann surface.
I tried to use the fundamental definition of Riemann surface. I think $f$ must be a holomorphic function, so we could find a collection of open sets s.t. $\cup _\alpha U_\alpha = X$, and we could holomorphically map the $U_\alpha$ onto open sets of $\mathbb C$
By Implicit function THM we could show that there exists a function $g$, such that $X$ coincides with the set of all points $(a,g(a))$. But how to prove that $(a,g(a))$ is locally homeomorphic to $\mathbb C$ everywhere?
Probably,
$$h:X\to\mathbb C$$ $$(a,g(a))\mapsto a$$ is a diffeomorphism from $X$ to $\mathbb C$.
But why do we need the slope of tangent lines to be non-zero?
Since we are only using one chart here, we do not have to prove that the transition map is holomorphic?
More a long comment than an answer.
First, is false that "exists a function $g$, such that $X$ coincides with the set of all points $(a,g(a))$". The condition only is true locally (in a "chunk" of $X$). And possibly other chunk of $X$ is the image set of a function $b\mapsto(g(b),b)$. Think in the simpler (and real) example of the unit circle $x^2 + y^2 = 1$.
Second, a Riemann surface is more than a set. You need a complex structure: a family of charts with holomorphic transition maps. The functions $(a,g(a))\mapsto a$ and $(g(b),b)\mapsto b$ are the obvious choice.