= Solution
{c}
Fix $x\in U$. Since $U$ is open, choose an open <ball> $B(x,\varepsilon)\subset U$. For any $y$ in this ball the segment $x+t(y-x)$ remains in it. For each coordinate $G_j$, the <chain rule> gives
$$
\frac d{dt}G_j(x+t(y-x))=DG_j{}_{x+t(y-x)}(y-x)=0.
$$
The ordinary <mean value theorem> therefore gives $G_j(y)=G_j(x)$ for every coordinate, hence $G(y)=G(x)$. Thus $G$ is a <locally constant function>.
For each $z\in\mathbb R^m$, its <preimage> is a union of such open balls, so $G^{-1}(\{z\})$ is open in $\mathbb R^n$. Its complement in $U$ is the union of the other fibers and is also open. If $U$ is nonempty and <connected>, the nonempty fiber through any chosen point is both open and closed in $U$, and must be all of $U$. Therefore
$$
\boxed{DG\equiv0\quad\Longrightarrow\quad G\text{ locally constant, and constant on connected }U.}
$$
The empty-domain case is vacuous. Notice that continuity of $DG$ was never needed.
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