$C^n$-manifold, which is not a $C^{n+1}$-manifold 
Question: Given an integer $n \in \mathbb{N}$, then what is an example of a $C^n$-manifold which is not a $C^{n+1}$-manifold?

Minor question: I know everything in theory, but I can not do it in practice. For instance, how can I check that the graph of the function $f\colon \mathbb R^2 \to \mathbb R^3$, $f(x,y)=(\sin x,e^y\cos x, xy)$, is a differentiable manifold.
Please do not use theorems, if it is possible. I want to see the whole procedure for at least one time. Any link and simple references are welcome.

I know the definition of a differentiable (or ~smooth or ~$C^{\infty}$) manifold, which is a Hausdorff, second countable topological space, with a differentiable atlas on it. If an atlas was given, then I feel comfortable with the definition of transition maps, and diffeomorphism (invertible smooth functions, with the smooth inverse), etc.
But I can not realize what is the Atlas in that example? Even if it was given to me, I don't have any idea how can I check all the infinitely many transition maps and diffeomorphism in practice.
 A: I interpret your question in the following sense:
A $C^n$-manifold $M$ is determined by a $C^n$-differentiable structure which we may regard as an atlas such that all transition maps are $C^n$ (let us call it $C^n$-atlas). Of course we could assume that the atlas is maximal with respect to this property, but it is irrelevant. Then the question is

Is every $C^n$-atlas also a $C^{n+1}$-atlas?

Variants of the question are

*

*If $M$ has a $C^n$-atlas, does it also have a $C^{n+1}$-atlas (which may be unrelated to the first atlas)?


*If $M$ has a $C^n$-atlas, does it always have a $C^{n+1}$-subatlas?
The answer to 2. (and a fortiori also to 1.) is "yes" if $n > 0$. See Every $\mathcal{C}^1$ manifold can be made smooth? For $n= 0$ the answer to 1. (and a fortiori also to 2.) is "no".  See Manifold that is NOT smooth
Let
$$h : \mathbb R \to \mathbb R, h(x) = \begin{cases} x & x \le 1 \\ x + (x-1)^{n+1} & x \ge 1\end{cases}$$

*

*$h$ is a homeorphism, thus $h^{-1} : \mathbb R \to \mathbb R$ exists.


*$h$ and $h^{-1}$ are $C^n$. This is trivial for $n = 0$. For $n > 0$ we get
$$h'(x) = \begin{cases} 1 & x \le 1 \\  1 + (n+1)(x - 1)^n & x \ge 1\end{cases}$$
$$h''(x) = \begin{cases} 0 & x \le 1 \\  (n+1)n (x - 1)^{n-1} & x \ge 1\end{cases}$$
$$h^{(n)}(x) = \begin{cases} 0 & x \le 1 \\   (n+1)!(x-1) & x \ge 1\end{cases}$$
Since $h'(x) > 0$  for all $x$, the inverse function theorem shows that $h^{-1}$ is $C^1$. But then $h^{-1}$ is even $C^n$. See the answer to Example of a $C^k$-homeomorphism $h : \mathbb R \to \mathbb R$ whose inverse is $C^{k-1}$, but not $C^k$


*$h$ is not $C^{n+1}$ because $h^{(n+1)}(1)$ does not exist.
Now let $\mathfrak A$ be the atlas on $\mathbb R$ having the two charts $id, h$. Transition functions are $h$ and $h^{-1}$ which are $C^n$. Thus $(\mathbb R, \mathfrak A)$ is a $C^n$-manifold. It is not a $C^{n+1}$-manifold by 3. However, the two subatlases $\{id\}$ and $\{h\}$ are $C^\infty$.
You can generalize this example to $\mathbb R^k$ by considering the map $h_k(x_1,\ldots,x_k) =(h(x_1), \ldots, h(x_k))$.
