What is some smooth parameterization of $y - |x| = 0$?

Question

Apparently, $\{(x,y) \in \Bbb R^2, y - |x| = 0\}$ is an example of a level curve $C: f(x,y) = c$, such that $f$ is not smooth, but $C$ admits a smooth parameterization.

What is some smooth parameterization of the the level curve $C: y - |x| = 0$?

I am not sure how such a construction is possible. We would like to have $\gamma: (\alpha, \beta) \subset \Bbb R \to \Bbb R^2$, such $\gamma(t) = (\gamma_1(t), \gamma_2(t))$ satisfies $\gamma_2(t) = |\gamma_1(t)|$, for all $t \in (\alpha, \beta)$. If $\gamma_1(t) \neq 0$ for all $t$, then $\gamma$ doesn't cover $C$. On the other hand, if $\gamma_1(t) = 0$ for some $t$, then $\gamma_2$ cannot be differentiable at this point. Am I missing something here?

Edit:

Where a function $f$ is said to be "smooth" if it is $C^{\infty}$

Answer 1

Reparametrize by $t^3$, that is, consider $\gamma(t) = (t^3, |t^3|)$. The issue is that this parametrization is smooth, but not regular, that is, we must have $\gamma'(t) = 0$ for some $t$ (note that $t=0$ does it, as expected).

We will never find a parametrization both smooth and regular.

Answer 2

I guess it's a liiittle too late, but I think I found a solution so I'll post it anyway.

The smooth parametrization is $$\gamma(t)=\left(e^{-\frac{1}{t^2}}\cdot t,\: e^{-\frac{1}{t^2}}\cdot |t|\right), \quad \gamma(0) = 0.$$

It's image is clearly all of $\{(x, y) \:\mid\: y = |x|\}$. It is obviously smooth everywhere except $t=0$, and at the only "problematic" point $t = 0$ all one-sided derivatives will exist and will be equal to zero, thus all derivatives are defined and are continuous, proving $\gamma$ to be $C^\infty$.