Let $\gamma (t) = (x(t), y(t))$ be a smooth regular plane curve $\gamma: I \to \mathbb R^2$ where $I$ is some open interval.
Now consider the following exercise:
Let $\varphi (u,v) = (x(u), v + y(u))$. Show that $\varphi $ is injective if and only if $x$ is. With $\gamma (t) = (t^3, t)$ show that $\varphi$ is bijective but $\varphi^{-1}$ is not smooth at points where $x=0$.
Show that for any $\gamma$, $\varphi$ is a local diffeomorphism at $(u_0,v_0)$ provided $x'(u_0)\neq 0$, that is, provided the tangent to the curve at $t=u_0$ is not vertical. If $x'(u_0)=0$ then $y(u_0)\neq 0$ (why?). In that case redefine $\varphi$ to be $\varphi (u,v)$ to be $(v + x(u), y(u))$: this is a local diffeomorphism at $(u_0,v_0)$.
My question is not about the solution, I think I can do that, but rather about the insight I should gain here. For example, why would I define such a map $\varphi$?
For example, $\gamma$ is a local diffeo iff $x'(t)\neq 0$ so saying the same about $\varphi$ does not seem to yield any new insight!