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In the statement of the implicit function theorem, we require the function $f$ to be $C^1$. Based on information found here, here, and here, I see that it is possible to relax the $C^1$ requirement and obtain a continuous form of the implicit function theorem. That is not my question.

What I'm interested in is this: does there exist $f:\mathbb{R}^2\to\mathbb{R}$, $f$ differentiable but not $C^1$, such that $f(0,0)=0$, $\frac{\partial f}{\partial y}(0,0)\neq 0$, yet there do not exist open set $U\ni 0$, $V\ni 0$, and $g:U\to V$ such that $f(x,g(x))=0$ for all $x\in U$ (i.e. there the is no implicit function)?

I'm having trouble constructing a counterexample. I got onto this question because, in the case of the inverse function theorem, if we relax $C^1$ to differentiable, we can use $f(x)=x+2x^2\sin(1/x)$ as a counterexample. I am looking for an analogous counterexample for the implicit function theorem.

Many thanks

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At some level, the implicit & inverse functions are essentially the same thing.

Hence choose $\phi(x,y) = f(x)-y$ and note that ${\partial \phi(0,0) \over \partial x} = {\partial f(0) \over \partial x}$.