Let $U = \{x \in \mathbb{R}^{m}; |x_i| < 1, i = 1, \ldots, m\}$ and $f: U \rightarrow \mathbb{R}$ a differentiable function such that $|\partial_{x_i} f(x)| \leq 3$ for all $x \in U$. Then, $f(U)$ is an interval of length $\leq 3m$.
Notice that $U = B(0, 1)$ with the $\operatorname{max}$ metric. Then, $U$ is a convex (and, thus, connected) set. Also, given that $f$ is differentiable, $f$ is continuous. Both observations guarantee that $f(U)$ is a connected set of $\mathbb{R}$. Therefore, $f(U)$ is an interval, since these are the only connected sets of the real line. Now, take two arbitrary points $x,y \in U$. By the Mean Value Theorem, there exists $\theta \in (0,1)$ such that $$|f(x) - f(y)| = \left|\sum_{1 \leq j \leq m}\partial_{x_j} f((1-\theta)x + \theta y)(y_j - x_j)\right|.$$ We can bound the latter by $6m$ using the triangle inequality. It seems that this is the best bound we can have. I don't see any straightforward way to get to $3m$. Is it even possible to do so?
Since $f(x) = 3\sum_{i=1}^m x_i$ satisfies the hypotheses and $f(U) = (-3m, 3m)$ is an interval of length $6m$, that's definitely the best you can do.
I expect there is a typo somewhere, but the simplest resolution is just replacing $3m$ with $6m$.