Let $F:D\subseteq\mathbb{R}^2\to\mathbb{R}$, $D$ open and connected set, be a $C^1 (D)$ application. 


What are the minimum requirements for $F$ such that the solutions of the equation $F(x,y)=0$ are given by $\gamma:I\subseteq\mathbb{R}\to\mathbb{R}^2$, $I$ open interval, $\gamma \in C^{1}(I),\ \gamma'(t)\neq 0,\ \forall t\in I$?


More precise, we want $F(x,y)=0\ \Leftrightarrow\ \exists t\in I, \gamma(t)=(x,y)$.



Regular curve given implicitly
It suffices that $\nabla F\neq0$ on the set $N=F^{-1}(0)$ and that $N$ is connected.
In this case you can use the implicit function theorem to describe $N$ as a graph of a $C^1$ function $\mathbb R\to\mathbb R$ in suitable rotated coordinates locally near every point.
This implies that near every point you can write $N$ as an image of a curve $\gamma$ that has $\gamma'\neq0$, and all that remains is to glue these pieces together.
You get a countable collection of curve segments and you can assume that at most two overlap at any point of $N$, so you should be able to build a $C^1$ curve $\gamma:\mathbb R\to D$ so that $\gamma(\mathbb R)=N$ and $\gamma'\neq0$ everywhere.


This assumption is not optimal, but I don't think there is an easy way to relax it.
If you allow $\nabla F$ to vanish on $x\in N$, the resulting curve $\gamma$ may need to branch or have a corner at $x$.
Consider, for example, $F(x)=x\_1x\_2$.
You can remove the connectedness assumption by considering several curves instead of one.
However, the condition $\nabla F|\_N\neq0$ is not necessary; see what happens when replacing $F$ with $F^2$.


