$G =graph(|x|)$. Show that exists a path $f:\mathbb{R} \to \mathbb{R}^{2}$, of class $C^{\infty}$, such that $f(\mathbb{R}) = G$ and $f(0) = (0,0)$ 
Let $G \subset \mathbb{R}^{2}$ the graphic of funtion $y = |x|$. Show that exists a path $f:\mathbb{R} \to \mathbb{R}^{2}$, of class $C^{\infty}$, such that $f(\mathbb{R}) = G$ and $f(0) = (0,0)$. Prove that all differentiable path $f$ under these conditions satisfies $f'(0) = 0$.

For the first part, my problem is in continuity. I can find some parametrizations, for $f(x) = |x|$, but none of them are continuous (I mean,$C^{n}$ for $n \geq 1$).
For the second part, since $f'(t)$ is continuous, so 
$$\lim_{t \to 0}f'(t) = f'(0).$$
But, I don't know how to proceed. Maybe, I should suppose $f'(0) \neq 0$, so $$\lim_{t \to 0}(\lambda_{1}'(t), \lambda_{2}'(t)) \neq 0,$$
then we can suppose that $\lim_{t \to 0}\lambda_{1}'(t) = c \neq 0.$

I appreciate any hint!
 A: Hint
Can you elaborate a strictly increasing function $f : [0,\infty) \to [0,\infty)$ such that $f^{(n)}(0)=0$ for all $n \ge 0$?
A way to do so is to tinker the map $x \mapsto e^{-\frac{1}{x^2}}$.
A: Consider 
$$h(t)=\begin{cases}0,&t\le 0\\e^{-1/t},&t>0\end{cases} $$
Then $h$ is $C^\infty$ (!) and strictly increasing on $[0,\infty)$, with $\lim_{t\to+\infty}h(t)=1$.
With this helper function, we can take
$$ f(t)=\bigl(t(h(t)-h(-t)), t(h(t)+h(-t))\bigr). $$

Let $f$ any such function. The idea is: If $f'(0)\ne 0$ then $f(t)\approx tf'(0)$ for $t\approx 0$, which is incompatible with the kink of the graph. 
More formally, we have
$$ \lim_{t\to0}\left|\frac1tf(t)-f'(0)\right|=0.$$
Note that $\frac1tf(t)\in G$, hence the above  $f'(0)$ cannot have positive distance to $G$. In other words, $f'(0)=(a,|a|)$ for some $a$.
Taking scalar products with $f'(0)$, we see that 
$$ \frac1t\langle f(t),f'(0)\rangle\to 2a^2$$
as $t\to 0$. But $\langle f(t),f'(0)\rangle \ge 0$ for all $t$ so that certainly $\frac1t\langle f(t),f(0)\rangle \le 0$ for $t<0$, so that for the limit we have $2a^2\le 0$, i.e., $a=0$.
