We need to find if x<y

I) if 2x <3y
2x<3y
=> x < 3/2y
If x=1, y=1 ,x < 3/2y, x=y
If x=1, y=10 ,x< 3/2y, x<y
So, we don't know if x=y or x<y. Insufficient

II) if xy > 0
Using the same cases as above, insufficient.

Together, using the same cases as above, still insufficient.

Hence E is the right answer.

Consider the case
1. 2x < 3y
x = 1, y = 1.
2x < 3y => 2 < 3 but x = y
x = 2, y = 3
2x < 3y => 4 < 9 => X < Y INSUFF

2. xy > 0
x = y = 1 => xy > 0 but x = y
x = 2, y = 3 => xy > 0 x < y INSUFF

Combining both we can have both the possibilities.

IMO E.

Great approaches above but we can speed up even further by replacing math with reasoning.

The question is asking you whether one thing is smaller (or bigger) than another.

(1) tells you that 2 of one thing is smaller than 3 of another. That's not enough to tell whether one thing is smaller (bigger) than the other. Insufficient.

(2) tells you that x and y share the same sign. Well, that tells you nothing about their relative magnitude; (2) is completely useless.

Combining, we have insufficient (1) with useless (2) so in combo we still have no sufficiency.

Choose (E).