Does the implicit function theorem hold for discontinuously differentiable functions? (This was posted on math.SE over 5 days ago and has not been answered,

although a comment mentioned a similar question on this site.)
Wikipedia's statement of the implicit function theorem requires that the original function

be continuously differentiable. $\:$ Is it known whether or not that condition can be removed?

If it can't be completely removed, can it be replaced with

"differentiable function whose derivative is bounded"?
 A: I haven't checked Terry Tao's proof of his inverse function theorem (Theorem 2 here), but if the proof (and hence the theorem) is correct, then from the theorem one gets the following implicit function theorem.
Let $\Omega\subset\mathbb R^n\times\mathbb R^m$ be an open set, and let $f:\Omega\to\mathbb R^m$ be an everywhere differentiable function, such that for all $(x_0,y_0)\in\Omega$ the partial derivative map $\partial_2 f(x_0,y_0):\mathbb R^m\to\mathbb R^m$ is invertible. Then for all $(x_0,y_0)\in\Omega$ with $f(x_0,y_0)=0$ there are an open neighbourhood $U$ of $x_0$ and an open neighbourhood $V$ of $y_0$ such that the set $U\times V\cap f^{-1}[\{0\}]$ is a continuous function $U\to V$ .
For the proof, one applies Tao's theorem to the map $f_0:\Omega\to\mathbb R^n\times\mathbb R^m$ defined by $(x,y)\mapsto(x,f(x,y))$ for which one first get the following result. Given any $(x_0,y_0)\in\Omega$ with $f(x_0,y_0)=0$ , there are an open neighbourhood $W$ of $(x_0,y_0)$ and an open set $W_1$ containing $(x_0,0)$ such that $f_0\,|\,W$ is a homeomorphism $W\to W_1$ . In particular, $f_1=(f_0\,|\,W)^{-1}$ is a continuous function $W_1\to W$ . Hence also the map $g_1:U_1=\{\,x:(x,0)\in W_1\,\}\to\mathbb R^m$ given by $x\mapsto{\rm pr}_2\circ f_1(x,0)$ is continuous. Noting that $\mathbb R^n\times\mathbb R^m$ has the product topology, one may suitably shrink the open sets $W$ to $U_2\times V$ and $U_1$ to $U$ so that by finally taking $g=g_1\,|\,U$ one by elementary set theoretic verifications shows the claim for $g=U\times V\cap f^{-1}[\{0\}]$ .
