Let $\mathbb{T}^n=\mathbb{R}^n / \mathbb{Z}^n$ and $u:\mathbb{R}\rightarrow \mathbb{R}$ be a $C^1$ function such that there exist $\omega_1, \dots, \omega_n \in \mathbb{Z}$ ($\mathbb{Z}$-linearly independent, such that $\{ (\omega_1x, \dots , \omega_n x)\in \mathbb{T}^n: x\in \mathbb{R}\}$ lies dense in $\mathbb{T}^n$) and a continuous function $f:\mathbb{T}^n \rightarrow \mathbb{R}$ such that
$$ u(x)= f(\omega_1 x , \dots, \omega_n x). $$
Is $f$ always $C^1$?
My intuition is that regularity of $u$ can only tell you about differentiability of $f$ in the $\omega = (\omega_1,\ldots,\omega_n)$ direction - at least this is all we get from the chain rule. Thus I'm going to try to construct a counterexample.
On some small ball you should be able to cook up a continuous but not differentiable function that is smooth when restricted to any line segment in the $\omega$ direction - for example, take any continuous but not differentiable function $g$ of $n-1$ variables and compose it with the orthogonal projection on to $\omega^\perp$, so that it is constant in the $\omega$ direction. Extending this to a continuous function $f: \mathbb T^n \to \mathbb R^n$ by multiplying by a smooth cutoff function and defining $u$ by your formula $u(x) = f(x\omega)$ we can verify that