Differentiable surface question Let $U$ be an open set in $\mathbb{R}^2$ and $F:U\to \mathbb{R}^3$ a one-to-one differentiable function such that its inverse from $F(U)$ to $\mathbb{R}^2$ is also continuous. Is it possible that $dF$, the derivative of $F$ as a matrix,  is not of rank 2 at some point of $U$?

Yes. Consider the function $F:\mathbb{R}^2\to\mathbb{R}^3$ defined by $F(x,y) = (x^3,y^3,0)$. The inverse of the function is $(a,b,0)\mapsto (a^{1/3},b^{1/3})$ which is continuous (but not differentiable at 0) and the rank of $dF$ at the origin is zero.
