Q:

Is every implicit function reparametrized?

Consider a continously differentiable non-constant function $f:\mathbb{R}^2\to\mathbb{R}$. Define
$$
K=\{x\in\mathbb{R}^2|f(x)=0\}.
$$ 
I wish to know whether there is a continuously differentiable function $g:\mathbb{R}\to\mathbb{R}^2$ such that 
$$
\forall k\in K,\quad\exists u\in\mathbb{R}, \quad g(u)=k.
$$
So it is like $K$ can be expressed as an image of a continuosly differentiable function $g$.
If $f$ is linear there of course we can find $g$. It is still true for any function? Any counter example?
I am thinking $f(x,y)=x^3+y^3-xy$. But, how would we prove it mathematically?

A:

Under your hypotheses, the answer is : no.
Basically because the existence of such function $g$ would mean that the zeroes of your function $f$ is a one-dimensional manifold, which is not true in general. 
Take for instance $f(x)=\text{dist}^2(x;C)$, where $C$ is the unit disc: $f$ is smooth, and here $K=C$. But you cannot express the unit disc as the image of a one-parameter continuously differentiable function. Eventually you could do it with a pathologic non-differentiable function.
More generally, as soon as $C$ is a closed convex set, the function $$f(x)=\text{dist}^2(x;C)$$
is a continuously differentiable function (having even a Lipschitz continuous gradient), whose zeroes are exactly  $C$. So basically proving your statement would prove in particular that you can parametrize any closed convex set with a differentiable function.

