(E) for me

(x-y)/(x+y) > 1 ?

<=> (x-y)/(x+y) - 1 > 0

<=> [(x-y) - (x+y)]/(x+y) > 0

<=> -2*y/(x+y) > 0

That implies

o -2*y > 0 thus y < 0 and x+y>0 thus x > -y > 0 (case 1)

or

o -2*y < 0 thus y > 0 and x+y<0 thus x < -y < 0 (case 2)

Stat 1
x > 0 is not enough. We have to know something about y such as y < 0.

INSUFF

Stat 2
y < 0 is not enough. We have to know something about x such as x > -y.

INSUFF

(1) and (2) together
As y < 0 and x > 0, We are more in the case 1. But we have to know if x > -y to conclude.

INSUFF