Is it possible to make the set $M:=\{(x,y)\in\mathbb{R}^2:y=|x|\}$ into a differentiable manifold? 1) Is it possible that be given a suitable smooth atlas to the set $M:=\{(x,y)\in\mathbb{R}^2: y=|x|\}$ so that $M$ to be a differentiable manifold? Why? How?
2) How can I prove that M is not an embedded smooth  submanifold of $\mathbb{R}^2$?
In really, How can I show that  $M$ is not the image of any $C^{\infty}$- immersion of $\mathbb{R}$ into $\mathbb{R}^2$? 
 A: Via transport of structure, any set with the same cardinality as $\mathbb{R}$ (such as your set $M$) can be given the structure of a smooth manifold in such a way that it is diffeomorphic to $\mathbb{R}$. 
In other words, simply choose a bijection $f:\mathbb{R}\to M$ (there's an obvious one in your case), then


*

*define a topology on $M$ by declaring that, for every open set $U\subseteq\mathbb{R}$, the set $f(U)$ will be an open set of the topology on $M$, then

*define an atlas on $M$ by declaring that, for every chart $(U,\psi:U\to V)$ of the usual atlas of $\mathbb{R}$, the chart $(f(U),(\psi\circ f^{-1}):f(U)\to V))$ will be an element of the atlas on $M$
In fact, in the case of your set $M$ (with its subspace topology from $\mathbb{R}^2$), the obvious bijection $f:\mathbb{R}\to M$ is already a homeomorphism, so there is no need to declare a new topology on $M$ as I described in my first step.
A: 
A differentiable manifold is a topological manifold equipped with an atlas whose transition maps are all differentiable.

Now by assuming that $M$ is a topological manifold, one can define an atlas on $M$ which have only one chart!
Consider chart $(M,\varphi)$ where its coordinate function, $\varphi$, is defined as follows:
$$\varphi:M\to \mathbb{R}$$
$$\varphi (x,|x|):=(x,0)$$
and its inverse is 
$$\varphi^{-1}:\mathbb{R}\to M$$
$$\varphi^{-1} (x)=(x,|x|)$$
easily you can show the map is homeomorphism. The only transition map is identity map that is smooth, clearly. So, $M$ is a smooth manifold.
A: In regards to your last question, by considering left and right-sided derivatives, show that the derivative of any such immersion at a point in the preimage of the origin must lie in the tangent space (at the origin) of the graph of line y= -x and in the tangent space  (at the origin) of the graph of the line y=x (both such tangent spaces thought of as linear subspaces of $T_{(0,0)}(\mathbb{R}^2$)). Since the intersection of these spaces is 0, the derivative cannot be injective here.
