If $\frac{\partial F^i}{\partial x^j}=0$ on a connected open set, is $F$ constant? Let $U$ be open in $\mathbb{R}^n$ and let
$$F:U\to \mathbb{R}^m$$
be a smooth map, i.e. $F\in C^\infty(U)$. It is easy to prove that if $U$ is convex and
$$\frac{\partial F^i}{\partial x^j}=0\tag{1}$$
everywhere on $U$ and for all $i,j$, then $F$ is constant on $U$.
But what if $U$ is only connected, but not necessarily convex? Do we have a smooth map $F$ such that $(1)$ holds, yet $F$ is not constant?
 A: $F$ will still be a constant map. Indeed, $U$ being a connected open subset of $\mathbb{R}^n$, it is path connected (here is a proof). Let $x_0,x_1\in U$ be arbitrary. There is a path $\gamma:[0,1]\to U$ such that $\gamma(0)=x_0$ and $\gamma(1)=x_1$. Now, $I$ is compact and $\gamma$ is continuous, so $\gamma(I)$ can be covered by finitely many open balls $B\subseteq U$ (which are convex). Then, by what you know about convex open sets, we easily get that $F(x_0)=F(x_1)$.
More Generally: One can prove that if $F:M\to N$ is any smooth map between differentiable manifolds (such as spheres, tori, etc.) and $M$ is connected, then $F$ is constant if and only if its differential $dF_p:T_pM\to T_{F(p)}N$ is the zero map for each $p\in M$ (which is exactly your equation $(1)$ in case $M\subseteq\mathbb{R}^n$ and $N\subseteq\mathbb{R}^m$ are open).
A: Any function on a connected topological space $X$ that is locally constant on $X$ is globally constant on $X.$* The open convex result mentioned above gives the locally constant property in this problem, and the connectedness of $U$ finishes it.
*The proof is a typical "the good set is nonempty and clopen, therefore it's everything" argument.
