Statement(1):
x + y > 0
This can be true for (x,y) = (4,-2) or (-2, 4)
INSUFFICIENT

Statement(2):
y^x < 0
Any positive number raised to any power is always positive.
For the above statement to hold true, y has to be negative (y < 0) and x has to be an odd integer.
If y = -3, x = -7, -5, -3, -1, 1, 3, 5,..
INSUFFICIENT

Let's combine.
If y < 0, and x + y > 0, then x > 0.
If x > 0 and y < 0, then x > y

[spoiler](C)[/spoiler] is the answer

If x and y are integers, is X > Y?

1) X + Y > 0

X > -Y (subtracting Y from both sides). It doesn't necessarily mean that X>Y. So, Statement I is insufficient to answer the question.

2) Y^X < 0

Y^X < 0, implies Y is a negative integer and X is an odd integer. It doesn't necessarily mean that X>Y because X can be a negative odd integer less than Y or it can be an odd integer greater than Y. So, Statement II is insufficient to answer the question.

Statement I and II

Y<0 and X>-Y
So, Y<0, -Y>0,and X>-Y.
i.e 0<Y<-Y<X. We can now confirm that X>Y. So, Statement I + II combination is sufficient to answer the question.

Answer C