I have $y'=f(y)$ and $f \in C^{\infty}(\mathbb{R})$
I want to show, that each solution is indefinitely often differentiable in its existence intervall.
Proof: Because of $f \in C^{\infty}(\mathbb{R})$ each solution is unique and exists. Therefore it holds $y'=f(y)$ and because of $f \in C^{\infty}(\mathbb{R})$ follows $y \in C^{\infty}(\mathbb{R})$ I hope, that's ok.
There is a second question to this: Solution is constant, if it has local max or min.
How can I proof that?
What you propose in your answer to the first question is circular reasoning, I am afraid: a composition of a $C^1$ and $C^{\infty}$ functions need not be of class $C^{\infty}$.
A proof should proceed by induction. $t \mapsto y(t)$ is clearly of class $C^1$. Assume that $t \mapsto y(t)$ is of class $C^k$, so $t \mapsto y'(t) = f(y(t))$ is $C^k$, too, but this means that $t \mapsto y(t)$ is of class $C^{k+1}$.
Regarding your second question, as $f$ is assumed to be $C^{\infty}$, the uniqueness theorem holds. Let $t_0 \in \mathbb{R}$ be such that a solution $y(\cdot)$ has a local extremum at $t_0$. Then $y'(t_0) = 0$. But observe that both $y(\cdot)$ and a function constantly equal to $y_0 := y(t_0)$ are solutions to the initial value problem \begin{equation*} \begin{cases} y' = f(y) \\ y(t_0) = y_0, \end{cases} \end{equation*} so they must be identical. (Indeed, without uniqueness the property described in your second question holds, too, but a proof is much more tricky.)