In what context would one think a cone surface is not a smooth manifold? Here the cone surface is $C:=\{z^2=x^2+y^2,\, z\ge 0\}$ in $\Bbb R^3$. Sometimes it is said $C$ is not a 2-dimensional smooth manifold. Talking about manifolds without specifying the atlas is meaningless, of course. However, it's pretty easy to see that under many atlases, for example any atlas that's compatible with the atlas containing the projection to the $xy$-plane as the only chart, $C$ will indeed be a smooth manifold. So I wonder, in what "usual" context (topology, atlas etc), would anyone possibly regard $C$ not as a smooth manifold?
 A: It is not a smooth manifold in the sense of the differentiable structure it inherited as a subspace of $\Bbb R^3$. For instance, the smooth function $f:\Bbb R^3\to\Bbb R$ given by $(x,y,z)\mapsto z$ is not differentiable as a function on the cone.
To elaborate a bit, a subset $S$ is an embedded $n$-dimensional submanifold of an $m$-dimensional manifold $M$ iff there for every $s\in S$ exists an open $U\subseteq M$ with $x\in U$ and a chart $\phi:U\to \Bbb R^m$ such that $\phi(U\cap S)$ is an $n$-dimensional hyperplane in $\Bbb R^m$. In that case, the different charts $(\phi|_{U\cap S}, U\cap S)$ may be extended to a maximal atlas on $S$, and this is the differential structure that $S$ inherits from $M$. There is no such thing for the tip of the cone in $\Bbb R^3$.
A: An elementary property (that is often needed) of smooth manifolds is the tangent space at each point, which is obtained by taking derivatives of curves. Consider any curve on $C$ that passes through the peak. This curve is not differentiable at the peak (you can consider it as a curve in $\mathbb{R}^3$), hence we are missing a tangent space at this point.
A: For surfaces in $\mathbb{R}^3$ there is also an equivalent geometrical property that some set is a surface (smooth submanifold of dimension 2, without border) if and only if locally it can be represented by a graph of a smooth function. It means, locally your surface has a form:
$$
 x = f(y,z) \text{ or } y = f(x,z) \text{ or } z = f(x,y),
 $$
where $f$ is some infinitely smooth function. This comes from the fact that 
if you have general parametrization $p(u,v) = (x(u,v), y(u,v), z(u,v)), \, (u,v)\in U,$
where $(u,v)$ is the parametrization, then the differential of $p$ is injective (by definition of a surface). Injectivity of $p$ will imply that matrix of the differential will have rank 2, which says that locally you parametrize your surfrace by some pair $(x,y)$ or $(y,z)$ or $(x,z)$. 
Then you need to work a little with the implicit function theorem to obtain 
the formula above. 
For the case of the cone it becomes obvious. The only way to choose coordinate system so that surface near the peak is a function, is to choose 
that peak "looks up". But in this case the obtained function is necessarily not smooth, hence it cannot be a surface. 
