ywilfred's explanation covers what I did but here's just detailed breakdown:
-from the stem, we know X and Y are positive so we only need to test pos numbers
(A) Y= 2X
-from this, we know Y is twice X. I used (x=1,y=2) and(x=2,y=4). You can select any number so long as it fits (A). (x=5,y=10) etc. But small numbers are easier to work with.
A) If x = 1, y = 2, then 4(1/9) < 1
If x = 2, y = 4, then 16(1/81) < 1
A is sufficient
Now the only possible choices are A and D.
-this says nothing about X so it's insufficient
Answer is A.
The logic you are following (two unknowns require two equations) doesn't apply here because we're not asked to find a specific value for either X or Y. We only need to know if
(4^X) (1/3)^Y < 1
And from (A), we know that the left side of this inequality will always be less than 1.