Everywhere differentiable inverse function theorem in which the derivative is invertible at only $1$ point

Question

I'm reading about inverse function theorem for everywhere (not necessarily continuously) differentiable funtions. First from Terence Tao's blog, i.e.,

Theorem 2 (Everywhere differentiable inverse function theorem) Let $\Omega \subset \mathbf{R}^n$ be an open set, and let $f: \Omega \rightarrow \mathbf{R}^n$ be an everywhere differentiable function, such that for every $x_0 \in \Omega$, the derivative map $D f\left(x_0\right): \mathbf{R}^n \rightarrow \mathbf{R}^n$ is invertible. Then $f$ is a local homeomorphism; thus, for every $x_0 \in \Omega$, there exists an open neighbourhood $U$ of $x_0$ and an open neighbourhood $V$ of $f\left(x_0\right)$ such that $f$ is a homeomorphism from $U$ to $V$.

The author fixes a point $x_0 \in \Omega$ and then proceeds to prove there is a local homeomorphism at $x_0$. In doing so, the author proves Lemma 5 whose proof appeals to the invertibility of $\partial f (x_1)$ for another $x_1 \neq x_0$, not just $\partial f (x_0)$.

I would like to ask if there is a stronger version of above theorem that requires only the invertibility of $\partial f (x_0)$, i.e.,

Let $\Omega \subset \mathbf{R}^n$ be an open set, and let $f: \Omega \rightarrow \mathbf{R}^n$ be an everywhere differentiable function, such that the derivative map $D f\left(x_0\right): \mathbf{R}^n \rightarrow \mathbf{R}^n$ of $f$ at $x_0 \in \Omega$ is invertible. Then $f$ is a local homeomorphism at $x_0$, i.e., there exists an open neighbourhood $U$ of $x_0$ and an open neighbourhood $V$ of $f\left(x_0\right)$ such that $f$ is a homeomorphism from $U$ to $V$.

Answer 1

This is false. As the inverse function theorem requires a function to be $C^1$, a natural candidate for a counterexample is a function with discontinuous derivative.

Let $f: x \in \mathbf{R} \mapsto x + x^2 \operatorname{sin}(1/x^2)$, where we set $f(0) = 0$. At $x = 0$ the derivative is $f'(0) = 1$, and everywhere else $f'(x) = 1 + 2x \operatorname{sin}(1/x^2) - 2/x\operatorname{cos}(1/x^2)$.

Now a local homeomorphism on $\mathbf{R}$ is strictly monotone, but $f'$ changes sign arbitrarily close to zero.