Let $f:I\to \mathbb R^2$, defined by $f(t)=(x(t),y(t))$ a differentiable path. I would like to prove the following implication
$$\text{If $x'(t)\neq 0$ for every $t\in I$, then $f(I)$ is a graph of a differentiable function $\xi:J\to \mathbb R$}$$
I'm trying to use the immersion theorem which is the following:
Theorem: Let $U\subset \mathbb R^m$ be an open set. If $f:U\to \mathbb R^{m+n}$ is an immersion, then there is a diffeomorphism $h$ such that $h\circ f=i$, where $i$ is the canonical inclusion.
We know that $f'(x):\mathbb R\to\mathbb R^2$ is injective for every $x$, since $f'(x)$ is non-zero for every $x\in U$ (due to the fact $x'(t)\neq 0$). Then we know there are open sets $U\in \mathbb R^2$ and $W\subset \mathbb R\times \mathbb R$ and a diffeomorphism $h:U\to W$ such that $h\circ f(x)=(x,0)$.
I'm almost there, I need help to finish this proof.