# Smooth curve with trace $\lvert x \rvert$

I want to prove the following:

(1) There is a parameter transformation from the continuous curve $c: \mathbb R \to \mathbb R^2; t \mapsto (t, \lvert t \rvert)$ into a smooth curve.
(2) Every continuous parameter transformation from $c$ into a smooth curve is not regular.

Attempt:
(1) I noticed that every bijective function $f$ that is smooth with $f^{(n)}(0) = 0$ works. Since polynomials become eventually constant with $f^{(k)} \neq 0$ for some $k\in \mathbb N$ we need some other function. From basic analysis we know that $f: \mathbb R \to \mathbb R$ defined by $$f(t) := \begin{cases}e^{-1/t} & t \neq 0 \\ 0 & t=0\end{cases}$$ is smooth with $f^{(n)} = 0$ for each $n\in \mathbb N$. But to make it bijective and satisfy $f(t) = -f(-t)$ we modify it:$$f(t) := \begin{cases}te^{-1/|t|} & t \neq 0 \\ 0 & t=0\end{cases}$$

It's easy to see that this function is indeed bijective and smooth. Finally, applying the transformation $f$ we get our smooth curve: $$d:\mathbb R \to \mathbb R^2; t \mapsto(f(t), \lvert f(t) \rvert)$$ Is this correct so far?
(2) Let $d = c \circ f$ be a continuous parameter transformation into a smooth curve. We need to show $\exists t\in \mathbb R$ such that $d'(t) = 0$. Claim: $d'(0) = 0$. But how do prove this? Using differential quotients I don't quiet get to what I want to show...

We have $d’=\bigl(f’,\frac{f}{|f|}\cdot f’\bigr)=f’\cdot(1,f/|f|)$. Now if $f’(0)$ wouldn’t be zero, $d$ wouldn’t be continuous in zero.