Given a function ${\bf f} (x,y,z) = (f_1, \dots, f_n) : \mathbb{R}^3 \to \mathbb{R}^n$, suppose I can write ${\bf f} (x,y,z) = ({\bf g} \circ p)(x,y,z)$ where $p(x,y,z) : \mathbb{R}^3 \to \mathbb{R}$ and ${\bf g}(t)=(g_1, \dots, g_n) : \mathbb{R} \to \mathbb{R}^n$ and ${\bf g}, p$ are "sufficiently nice". Then the Jacobian of $\bf f$, which is denoted by ${\bf J}_{{\bf f}}$, is rank-$1$ everywhere, since
$$ \mathbf{J}_{{\bf f}} (x,y,z) = \begin{bmatrix} \frac{d{\bf g}_1}{dt} \\ \frac{d {\bf g}_2}{d t} \\ \vdots \\ \frac{d {\bf g}_n}{d t} \end{bmatrix} \begin{bmatrix} \frac{\partial p}{\partial x} \\ \frac{\partial p}{\partial y} \\ \frac{\partial p}{\partial z} \end{bmatrix}^\top$$
Is the converse true? i.e. if the Jacobian $\mathbf{J}_{{\bf f}}$ is rank-$1$ everywhere, then does there always exist a decomposition ${\bf f} = {\bf g} \circ p$ as above? I am ok if we need to make mild assumptions on the functions ${\bf f}, {\bf g}, p$. I am not looking for pathological examples.
One idea I had was to use facts about constant rank maps between manifolds but I couldn't figure this out...