$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).

Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this [MathSE post](https://math.stackexchange.com/questions/139883/why-does-a-lipschitz-function-f-mathbbrd-to-mathbbrd-map-measure-zero-s) (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).