Let $X$ be the subspace of $\mathbb{R}^2$ consisting of the graph of the absolute value function. That is, $X=\{(x,|x|) : x\in\mathbb{R})\}$.
We define a functional structure on $X$ by restricting smooth real valued functions defined on open subsets of the plane: an open set $U\subset X$ has the form $U=V\cap X$, where $V\subset\mathbb{R}^{2}$ is open. Then define $f\in F(U)$ iff $f=g|_{U}$, where $g$ is a smooth real value function on $V$.
E.g., The function $\pi_{2}(x,y)=y$ is smooth on any neighborhood $V$ of the origin. Therefore, its restriction to $U=V\cap X$ is in $F(U)$ (note that the restriction is not smooth at the origin).
Question: Is $\mathbf{X}$ with the functional structure $\mathbf{F}$ a differentiable manifold? (Defining smooth manifolds via functional structures: here).
It seems to me that it is not a differentiable manifold. I tried showing that at the origin it is not locally isomorphic to $(\mathbb{R},C^{\infty})$ (the real line with the functional structure of smooth real valued functions). Assuming that there is a local isomorphism at the origin I was trying to get to a contradiction more or less using functions such as $\pi_{2}$ above, which are not smooth when restricted to $X$. Nothing has worked so far...
Note: Clearly $X$ is not a manifold in $\mathbb{R}^{2}$. This is discussed for example in this and this posts.