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Let $U$ be an open subset of $mathbbR^n$.
Suppose moreover that $U$ is $C^k$. ($0 leq k leq infty$).

This means that for each $x in partial U$, there exists an $r>0$ and a $C^k$ function defined on $mathbbR^n-1$ so that $B(x,r) cap U = B(x,r) cap (x,y):y>f(x)$ and $B(x,r) cap partial U = B(x,r) cap (x,y):y=f(x) $. You may "change the order of the coordinates" if needed.

Now, let $h:mathbbR^n rightarrow mathbbR^n$ be a $C^k$ diffeomorphism.

Question: Is $h(U)$ also a $C^k$ open set?

Remark: Clearly, $partial U$ is an embedded $C^k$ submanifold of $mathbbR^n$ so that its image $h(partial U)$ is also an embedded $C^k$ submanifold. Please note that this fact does not immediately solve the whole problem.

Any help will be fully appreciated.

This is true. The difficulty in proving it seems to be to arrange for the correct ball radius $r$ in the ball $B(x,r)$. However, that difficulty can be sidestepped by rewording the definition of a $C^k$ open set using "open neighborhoods" instead of "open balls".

$U subset mathbb R^n$ is a $C^k$ open set if and only if for each $x in partial U$ there exists an open neighborhood $V subset mathbb R^n$ of $x$ and a $C^k$ function $f$ defined on $mathbb R^n-1$, so that [everything holds as written except with $B(x,r)$ replaced by $V$].

The "open ball" definition clearly implies the "open neighborhood" definition.

Conversely, suppose that the "open neighborhood" definition holds. Then for each $x$ with corresponding open neighborhood $V$, we simply choose $r>0$ so that $B(x,r) subset V$, and now the "open ball" definition holds.

Now, assuming that $U$ is a $C^k$ open set and $h$ is a $C^k$ diffeomorphism, let's prove that $h(U)$ is a $C^k$ open set.

For each $x in partial(h(U))$, consider $h^-1(x) in partial U$. Choose a function $f$ and an open neighborhood $V$ which witness that $U$ is a $C^k$ open set near the point $h^-1(x)$. Then the function $f h^-1$ and the open neighborhood $h(V)$ witnesses that $h(U)$ is a $C^k$ open set near $x$.