Your idea for the reflexivity of the relation is good

^

which is obvious if we let . which will be in Q

Symmetric:

^ ^

since multiplication in Q is communative, i.e., .

If and , since both x and xy are rational then it follows there must be a y in Q such that . Apply the same argument but for x and you show

Transitive:

if and then

by , we know x and xy are in Q, and by we know y and yz are in Q. So we want to know if xz is in Q. It follows if y and yz are in Q then z must be in Q, by the same argument for the Symmetric. So if x is in Q and z is in Q, then xz must be in Q.