For $a, b \in \mathbb{R}$, consider the map \begin{eqnarray*} f_{a,b} : &\mathbb{R}^2 &\longrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathbb{R} \\ &(x,y) &\mapsto x^3 - (ax + b) - y^2 \end{eqnarray*}
I'm trying to find out for which values of $a$ and $b$, $S_{a,b} := f_{a,b}^{-1}(\{0\})$ is a $k$-dimensional submanifold of $\mathbb{R}^2$. This demands that for every point $p = (p_1, p_2) \in S_{a,b}$ there exist a $U_p \subseteq \mathbb{R}^2$ with $p \in U_p$ which is open in $\mathbb{R}^2$, and a differentiable map $f_p : U_p \rightarrow \mathbb{R}^{2-k}$, such that $$f_p^{-1}(\{0\}) = U_p \cap S_{a,b} \,\,\, \wedge \,\, \text{rk} f'(p) = 2-k.$$
Now if we take, for every $p \in S_{a,b}$, $U_p = \mathbb{R}^2$, then this significantly simplifies this. Namely, we can then simply take $k=1$ and $f_p = f_{a,b}|_{U_p}$, so that the first condition ($f_p^{-1}(\{0\}) = U_p \cap S_{a,b} $) is immediately satisfied. That leaves the second condition. Note that:
$$1 = \text{rk} f'(p) = \text{rk} [3p_1^2 -a \,\,\,\,\,\,\,\, -2p_2] = \text{dim sp}\{3p_1^2 -a, -2p_2\}. $$
The problem is that I don't see why this would not be the case for any $a,b$... Regardless the value of $a$ for example, we can still multiply $-2p_2$ by all real numbers to obtain the whole of $\mathbb{R}$ (unless $p_2 = 0$).
It's a calculation, followed by an application of the regular level set theorem. You have
$f_{a,b}(x,y)= x^3 - (ax + b) - y^2,\ \partial f_x(x,y)=3x^2-a$ and $\partial f_y(x,y)=-2y.$
First, suppose $a\ge0.$ Then, the critical points for $f$ are $\left(\pm\sqrt{\frac{a}{3}},0\right)$. Now, substitute these points into $f$ and set to $0.$
$\left(\pm\sqrt{\frac{a}{3}}\right)^3-a\left(\pm\sqrt{\frac{a}{3}}\right)-b=0.$ from which $b=\mp 2\left(\sqrt{\frac{a}{3}}\right)^3.$
Thus, on the complement of the set defined by the above equation, $f_{a,b}^{-1}(0)$ is a submanifold, (trivially) of rank $1$.
If $a<0$, there are no critical points, so $f_{a,b}^{-1}(0)$ is a submanifold for any value of $b$.