Let $f:E\mapsto E$ be a map such that $f\circ f\circ f = f$.
Show that: $$f \text{ injective} \Leftrightarrow f \text{ surjective}$$
Let $x\in E$, then $f(f(f(x)))=f(x)$.
We have: $$f \text{ injective} \Leftrightarrow \forall x\in E,\;f(f(x))=x$$ $$ \Leftrightarrow \forall x\in E,\;x=f(z) \text{ and } z=f(x)$$ $$ \Leftrightarrow f \text{ surjective}$$.
Do you validate this answer ?
Yes, your proof is broadly correct. A few notes, though:
Clearly $f$ being injective implies that $f^2$ is the identity. The reverse implication is less obvious, so you might want to flesh it out a bit.
Your second equivalence makes things less easy to understand, so I'd leave it out, and instead just note that (in the "forward" direction) we have, for every $x$, produced a $y = f(x) \in E$ such that $f(y) = x$ (using words makes this much clearer. Use words, words are useful. Humans can read them), which clearly implies surjectivity.
Why surjectivity of $f$ implies your third statement is entirely unclear. Indeed, I can't find a good way to prove it that doesn't essentially replicate a proof of the whole statement. Instead, note that if $f$ is surjective, then for every $x\in E$, there is a $y\in E$ such that $f(y) = x$. But then applying $f^2$ to both sides gives us $x = f(y) = f^3(y) = f^2(x)$, so we have your second statement, and hence that $f$ is injective.