I am really confused with graph of $|x|$ has a differentiable structure.
I know that if $X$ is a differentiable manifold homeomorphic to $Y$ then $Y$ gains differential structre from $X$. But I dont understand why at the singular point these differential structre doesnot fail?
As you say, if you have a differentiable manifold $X$ with differentiable structure $\mathcal D_X$ and a homeomorphism $h : X \to Y$ to a topological space $Y$, then $Y$ is a topological manifold which can be given a unique differentiable $h_*(\mathcal D_X)$ structure such that $h$ becomes a diffeomorphism. Note if $Y$ already has a natural differentiable structure $\mathcal D_Y$,then in general $h_*(\mathcal D_X) \ne \mathcal D_Y$. In fact $h_*(\mathcal D_X) \ne \mathcal D_Y$ iff $h : (X,\mathcal D_X) \to (Y,\mathcal D_Y)$ is a diffeomorphism. As an example consider $X = Y = \mathbb R$ with the standard differentiable structure $\mathcal D_\mathbb R$ and $h(x) = x^3$. This is a smooth map, but $h^{-1}$ is not. Thus $h$ is no diffeomorphism and $h_*(\mathcal D_\mathbb R) \ne \mathcal D_\mathbb R$.
I think what may be confusing in your example of the graph $G$ of $\lvert x \rvert$ is that $G$ is not a submanifold of $\mathbb R^2$ (the corner at $x = 0$ prevents this). But this does not prevent us from giving $G$ a differentiable structure. With such a differentiable structure the inclusion $i : G \to \mathbb R^2$ is not smooth.