Krdytkyu

Let R and S be two equivalence relation on X.

I wanna prove that $Rcirc S$ is an equivalence relation, but I can't prove that it is reflexive and transitive.

For transitivity:

There are arbitrary $x$ and $y$ with $(x,y)in Rcirc S iff$ there is $z$ with $(x,z)in R$ and $(z,y)in S$ ...?

And where I have to use the $Rcirc S=Scirc R$?

I use the notation $xRy$ instead of $(x,y)in R$. In particular, $xRSy$ means $xRaSy$ for some $a$. It is also convenient to write $xRySz$ instead of ($xRy$ and $yRz$). In general I would advise you do draw such relations as graphs, with vertices representing points, and edges denoting the relationship.

Assumption $Rcirc S=Scirc R$.

Transitivity: Suppose $xRSy$ and $yRSz$, i.e. $xRaSy$ and $yRbSz$ for some $a,b$.
Then $aSyRb$, or short, $aSRb$. Since $SR=RS$, we get $aRSb$, let's say $aRcSb$ for some $c$. But then
$xRaRcSbSz$, and since $R,S$ are equivalence relations, $xRcSz$, so $xRSz$.

I am confident that you can now prove reflexivity by yourself.

Assumption $Rcirc S$ is an equivalence relation.

This direction is very similar to the proof above, you should give it a try.