The function $R(x)=rank(Df(x))$ is locally constant on $\Omega$, i.e. it is constant in a neighbourhood of every point $x \in \Omega$. Let $f:\Bbb R^n \to \Bbb R^n$ be a mapping of class $C^1$. Prove that there is an open and dense set $\Omega \subseteq \Bbb R^n$ such that the function $R(x)=rank(Df(x))$ is locally constant on $\Omega$, i.e. it is constant in a neighbourhood of every point $x \in \Omega$.
I was first thinking to use rank theorem which states that if $Df(x_0)=m$ then $\exists $ a diffeomorphism $\Phi$ in the neighbourhood of $x_0$ and $\Psi$ in the neighbourhood of $f(x_0)$ such that $\Psi \circ f \circ \Phi^{-1}(x_1, \cdots , x_n)=(x_1, , \cdots , x_m)$.
Then I was thinking to take differentiation and then chain rule but it won't work as we will get $D(f( \Phi^{-1}))$.
The second way I was thinking was to use the dense property e.g:
Can we say anything about $\{x \in \Bbb R^n| D(f(x))\geq k\}$ and then work with it?
I deleted my earlier post as that was incorrect.
Please help..
 A: Sorry i misread your question. Maybe i can argue for $n<3$ by a modest argument below. I try a bit for $n\geq 3$ but i cannot generalize the argument. Also this is not a complete answer so it is suit only as a comment but it is too long apparently. I'm a novice to this subject so please take this with a grain of salt. Let me know if you found a satisfying argument.

Suppose $f \in C^1(\mathbb{R}^n,\mathbb{R}^n)$ for $n=1,2$. Let $\Omega= \{ x \mid R \text{ locally constant at }x \}$. By definition $\Omega \subseteq \mathbb{R}^n$ is open. We want to argue that $\Omega$ dense. Suppose not, then there is a point $x \in \Omega^c$ and a neighbourhood $U_x$ of $x$ such that $U_x \subseteq \Omega^c$. Means that $R$ is not locally constant at every point in $U_x$. By inverse function theorem, any regular point is in $\Omega$ so for any $y\in U_x$, we have $0 \leq R(y)<n$. 
For $n=1$, this means $R(y) = 0$ for all $y \in U_x \subseteq \Omega^c$. Contradict to the definition of $U_x$. For $n=2$, the value $R(y)$ for $y \in U_x$ is either $0$ or $1$. There must be a point $y_0$ in $U_x$ where $R(y_0)=1$ otherwise we have the same contradiction as above. Now by definition of $U_x$, $R$ is not locally constant at $y_0$. This means every neighbourhood $U_n$ of $y_0$ contain a point $y_n$ with $R(y_n)=0$. So we have a sequence $(y_n)$ with $R(y_n)=0$ converging to the point $y_0$ with $R(y_0)=1$. By continuity of partial derivatives of $f$ this is cannot happen.    
