First, we can simplify the left:

 [(x^-1)(y^-1)]^-1 = [  (xy)^-1 ] ^-1 = xy

(1) If x=2y:

Then xy = 2y²; also [ 1/x + 1/y ]^-1 = [ (1/2) (1/y) + 1/y ] ^-1 = 2y/3

Is 2y/3 > 2y² ? We can't know, because it is false if y is negative, but true if y is positive.

(2) If x + y > 0:

We can rewrite the left as [ (x+y)/(xy) ]^-1 = (xy)/(x + y).

Is (xy) / (x + y) > xy? Since x + y > 0, (x + y)^-1 is always between 0 and 1, and the statement is true.

Therefore (1) is not sufficient, but (2) is sufficient.