If $f$ is even and $y'=f(y)$ then $y$ is odd Let $f\in C^1(\mathbb{R}, \mathbb{R})$ be an even function.
Consider the maximal solution $y\colon\left]\alpha ,\beta\right[\to \mathbb{R}$ of the IVP $$y'=f(y),\ y(0)=0$$
Prove that $y$ is an odd function and $\beta =-\alpha$.

To be able to prove that $y$ is odd, I first need its domain to be symmetric ($x\in \left]\alpha ,\beta\right[\implies -x\in \left]\alpha ,\beta\right[$), from here I can conclude that $\alpha=-\beta$. But how to prove the domain is symmetric?
And how to prove that $y(-x)=-y(x)$ for all $x\in \left]\alpha,\beta\right[$? I suspect it has something to do with the fact that the derivative of an even function is odd and $y'=f(y)$, but I can't see how to get the desired result.
Clearly, $\alpha&lt;0&lt;\beta$.
Let $r=\min\{-\alpha,\beta\}$ so that we at least have a solution on $\left]-r,r\right[$.

Then show: If $y$ is a solution on $]\alpha,\beta[$, then $x\mapsto-y(-x)$ is a solution on $\left]-\beta,-\alpha\right[$.

By uniqueness, these coincide on $\left]-r,r\right[$.

Then you can combine $y$ with $x\mapsto- y(-x)$ to find a solution on $\left]-\max\{-\alpha,\beta\},\max\{-\alpha,\beta\}\right[$. By maximality, we conclude $\alpha=-\max\{\alpha,\beta\}=-\beta$.
I'll leave this here, for someone that has a similar question.

Since f is odd, y' = f(y)=f(-y) — This means that not only the function is odd, but f[y(t)]=f[-y(t)] (we'll get to this later)

From this, we come to the point that y(t) has a known point (0,0) , from the inicial condition. 

Let us consider a point α : 

f[y(α)] = y'(α) = f[-y(α)]

Considering now α and -α=β, we come to : 

f[y(α)] = y' = f[-y(-α)] =f[-y(β)]

This means that f[y(α)] = f[-y(β)] = f[-y(-α)]

This implies that y(α) = -y(-α) , condition for a odd function f(x) = -f(-x) , so y(t) is odd, and it's done.