Showing that a group $G$ such that 3 does not divide $|G|$ is Abelian.

I asked this question here Understanding the hint of a question to show that $G$ is Abelian. but I did not receive answers to all my questions. So I am attaching a trial to the solution of the question which I found online and could not fully understand it.

First

Here is the main question: Let $G$ be a finite group such that 3 does not divide $|G|$ and such that the identity $(xy)^3 = x^3 y^3$ holds for all $x,y in G.$ Show that $G$ is abelian.

And here is the hint I got for the question:

First show that the map $G rightarrow G$ given by $x mapsto x^3$ is bijective. Then show that $x^2 in Z(G)$ for all $x in G.$

And here is a trial for the solution:

My questions are:

1-should we show that the map $G rightarrow G$ given by $x mapsto x^3$ is bijective because of the following step in the solution above:

$$(y^2x^2)^3 = (x^2y^2)^3 implies (y^2x^2) = (x^2y^2)$$ because there is no element in $G$ with order $3.$?

2- Also, how the given solution showed that $x^2 in Z(G)$ for all $x in G$? And can this statement be restated as $x^2 in Z(G)$ for all $y in G$?

3- Is the given trial of the solution correct? if not, how can we correct it?

Could anyone help me in answering those questions, please?