I am wondering if there are any function(s) such that the conclusion of the implicit function theorem holds true, despite the function(s) not satisfying the conditions stated in the hypothesis. For quick reference, https://en.wikipedia.org/wiki/Implicit_function_theorem.
I've thought about it for awhile, tried to play around with some possibilities, and have come up with none. However, I believe I saw this question on an old exam.
The function $f(x,y)=x-y^3$ is a basic case. It's partial derivative $\frac{\partial f}{\partial y}=3y^2$ is zero at $(x,y)=0$, but the function has a local inverse there.
The implicit function theorem is a generalization of the case when $m=0,n=1$, where, if $f:\mathbb R\to\mathbb R$, and $f'(a)\neq 0$, then $f$ is $1-1$ on some neighborhood of $a$. Then $f(x)=x^3$ is an example where $f'(0)=0$ but $f$ is still $1-1$ on all of $\mathbb R$, and hence a neighborhood of $0$.
So the implicit function theorem is not meant to be necessary. The "problem points" are spotted with differentiation, but you can easily find examples where the problem points still don't break the local inverse.