Let $U \subset \Bbb R^n$ be open. What is a geometric characterization of the smooth real-valued functions $f$ on $U$ satisfying the condition $$df \wedge dx^1 = 0 ?$$
I've expanded the expression according to the definition of $df$, and am not sure my solution is correct.
$f$ is defined on an arbitrary open subset $U$ of $R^n$.
$$df = \sum_{i =1} ^{n} \frac{\partial f}{\partial x^i}dx^i$$
Thus: $$\begin{align}df \wedge dx^1 &= \sum_{i =1} ^{n} \frac{\partial f}{\partial x^i}dx^i \wedge dx^1\\&= \sum_{i =2} ^{n} \frac{\partial f}{\partial x^i}dx^i \wedge dx^1 .\end{align}$$
Thus for our given expression to be true, we require $\frac{\partial f}{\partial x^i}dx^i \wedge dx^1 = 0$ for all $i \neq 1$. Now, for these $i,$ we know that $dx^i \wedge dx^1 \neq 0.$ Thus these must be smooth functions that are constant along each $x^{i \neq 1}$ axis.
Am I wrong anywhere?
You are correct that the condition is equivalent to the p.d.e. system $$\frac{\partial f}{\partial x^2} = \cdots = \frac{\partial f}{\partial x^n} = 0,$$ but your interpretation is not quite correct.
Each equation $$\frac{\partial f}{\partial x^i} = 0$$ implies that $f$ is constant along any line segment contained in $U$ parallel to any of the $x^i$ axes, $i = 2, \ldots, n$, and this statement is equivalent to the system. Putting these statements together gives the equivalent conclusion that for every hyperplane $H_y := \{x^1 = y\}$, $f$ is constant on each component of $U \cap H_y$.
NB this is not the same as $f$ depending only on $x^1$---that is, that $f(x^1, \ldots, x^n) = g(x^1)$ for some function $g$ on $U$---even for connected $U$. In particular, we cannot, as claimed in the question, conclude that $f$ is constant on the $x^i$-axes, $i \neq 1$.
On the other hand, the system does imply that about each point in $U$ there is a neighborhood $V$ such that $f\vert_V$ depends only on $x^1$, and again, this statement is equivalent to the system.