Let $U$ be an open connected subset of $\Bbb R^n$ and $f : U \to \Bbb R$ be a differentiable function such that $D f (p) = 0$ for all $p \in U$ then $f$ is a constant function.
If we can prove that $f$ is constant about each point in $U$ i.e., let if $x \in U$, there exists $B(x, r_x) \subset U$. Then if we can show that $f$ is constant on the nbd $B(x, r_x)$ then we are done. I think we have to use MVT.
Help Needed!
Pick any two points $u$ and $v$ from $U$, and apply the MVT to the segment $[u,v]$ that connects them: this is possible provided that $u$ and $v$ are sufficiently close, since $U$ is open and open balls in $\mathbb{R}^n$ are convex. You will find in the end that $f(u)=f(v)$.
You can conclude in a standard way.
Pick $x_0 \in U$ and let $C=\{x \in U| f(x) = f(x_0)\}$. It is easy to see that $C$ is closed since $f$ is continuous. Since $U$ is open, you can use the mean value theorem to show that $C$ is open. Since $U$ is connected we have $C = U$.
