Let $X : \mathbf R^2 \to \mathbf R^2$ be a no-zero smooth vector field. I want to show (without background about vector bundles or manifolds, just if possible differentiable calculus in $\mathbf R^n$) that there is a smooth function $\psi : \mathbf R^2 \to \mathbf R^2$ such that $ d_p\psi(X_p)$ is constant $\forall\ p \in \mathbf R^2$.
Thanks in advance!
Interesting question.
If such a function $\psi$ exists then $\psi$ takes the integral curves of $X$ to a family of parallel lines in $\mathbb{R}^2$. This puts strong restrictions on the family of integral curves. For instance, the orbit space of $X$ - meaning the quotient space obtained by collapsing each integral curve to a point - must be homeomorphic to $\mathbb{R}$.
However, there are counterexamples where the orbit space of $X$ is not homeomorphic to $\mathbb{R}$. For example, you can put the universal cover of this foliation in each horizontal strip $\mathbb{R} \times [n-1,n]$. The resulting orbit space is not even Hausdorff, let alone homeomorphic to $\mathbb{R}$.