This is a little lenghty, and there's probably an shorter way to see this, but I want to prove the following: Let $f:N^n\to M^m$ be a smooth map from between smooth manifold and $\{U,x\}$, $\{V,y\}$ be charts on $M$ and $N$ respectively with $f(U)\subseteq V$. I want to show that if the derivative (push-forward) $Df=f_*$ vanish identically on $U$, then $f$ is locally constant on $U$.
My thought is to use the following theorem from Browder:
Let $A$ be a convex subset of $\mathbb{R}^n$ and $f:A\to\mathbb{R}^m$. If $f$ is differentiable at each point of $A$ and if $||Df_p||\leq M$ for every $p\in A$, then $||f(b)-f(a)||\leq M||b-a||$ for all $a,b\in A$.
So if figured since $y\circ f\circ x^{-1}$ is smooth map between the open sets $x(U)\subseteq\mathbb{R}^n$ and $y(U)\subseteq\mathbb{R}^m$ and $(y\circ f\circ x^{-1})_*=y_*\circ f_*\circ x^{-1}_*$, we see $f_*$ vanishes at $p\iff y_*\circ f_*\circ x^{-1}_*$ vanishes at $x(p)$. Suppose $f_*$ vanishes on $U\implies y_*\circ f_*\circ x^{-1}_*$ vanishes on $x(U)$ and $q\in x(U)$. Since $x(U)$ is open, take a small open ball $q\in B_q\subset x(U)$. Then for any points $a,b\in x(U)$, we can connect them with a straight segment $[a,b]$ (since $B_q$ is convex). Then since $y_*\circ f_*\circ x^{-1}_*=0<M$ for all positive $M$, by the theorem above we have $||y\circ f\circ x^{-1}(a)-y\circ f\circ x^{-1}(b)||=0\implies y\circ f\circ x^{-1}$ is constant on $B_q\implies f$ is constant on $x^{-1}(B_q)\subseteq U$. That is, $f$ is locally constant on $U$.
Does this look right?
That works, but it's working too hard...
Locally you have a map $f: \Bbb R^n \to \Bbb R^m$ whose Jacobian vanishes. So in particular the partials of all of the component functions vanish. Thus it's constant on lines (because if a function $\Bbb R \to \Bbb R$ has zero derivative, then it's constant), by noting that the directional derivatives can be written in terms of the partials by the chain rule. So, connecting any two points in $\Bbb R^n$ by a line, the function is constant.