How do you show that on Zx(Z\{0}) the relation (j,k)II(m,n) if jn=km defines an equivalence relation...?

For any (j,k) in Zx(Z\{0}) one has jk = kj and hence (j,k) || (j,k). This shows that || is reflexive.

If (j,k), (m,n), and (p,q) in Zx(Z\{0}) satisfy (j,k) || (m,n) and (m,n) || (p,q) then from the definition of || we conclude that (1) jn = km and (2) mq = np. Multiplying (1) by q we deduce that jnq = kmq; from (2) we deduce that jnq = kmq = k(np). Since n is nonzero we deduce from jnq = knp that jq = kp, i.e., that (k,k) || (p,q). This shows that || is transitive.

If (j,k) and (m,n) in ZxZ\{0} satisfy (j,k) || (m,n) then jn = km, from which we deduce that mk = jn, from which we deduce that (m,n) || (j,k). This shows that || is symmetric.

Since || is symmetric and transitive and reflexive it is an equivalence relation on Zx(Z\{0}).

A geometric way of looking at this is to consider elements of Zx(Z\{0}) as points in the Cartesian plane. If (j,k) and (m,n) are in Zx(Z\{0}), then (j,k) || (m,n) if and only if jn = km, if and only if j/k = m/n, if and only if the slope of the line from (0,0) to (j,k) is equal to the slope of the line from (0,0) to (m,n), if and only if (j,k) and (m,n) lie on the same line through the origin.

If you use the same symbolic recipe to define a relation || on ZxZ, the resulting relation is not an equivalence relation on ZxZ. It turns out to be reflexive and symmetric, but it is not transitive. For example, it is true that (1,0) || (0,0) (because 1*0 = 0*0) and that (0,0) || (0,1) (because 0*1=0*0), but it is not true that (1,0) || (0,1) (because 1*1 is not the same as 0*0). I hope this helped.