The implicit function theorem (let's just say 2-variable functions for now) says that at points at which $\frac{\partial f}{\partial y}$ is non-zero (equiv. the graph doesn't have a vertical tangent line), we can express $y$ as a function of $x$ locally around that point. The importance of this "vertical tangent" condition on the 1st derivative is illustrated in the example of a circle, because any neighborhood of the rightmost or leftmost points of the circle $y$ can not be written as a single-valued of $x$. However, the 1st derivative condition is not necessary, since for example the function $x^{1/3}$ has a vertical tangent at $x=0$, but we can still write $y=x^3$.
EDIT: user Peter Morfe has pointed out that this example does not work, so here is one that does work: $f(x,y) = x^2+y^3-1$ (graph $f(x,y) = 0$ on Desmos if you want to look at it) has $\frac{\partial f}{\partial y} = 3y^2 = 0$ at $(x,y) = (1,0), (-1,0)$, but it is clear that one can write $y$ as a function of $x$ around those points.
My question: is there a further condition on the 2nd derivative (since that deals with curvature) giving us more information about points at which the 1st derivative (partial w.r.t $y$) is $0$ that can be used to prove a stronger version of the implicit function theorem? And of course, if there is a such an extension for the 2-variable case, is there a general extension for arbitrary numbers of variables?