Solution
This is being solved assuming the second statement is xy(z^2) > 1
The question is Is xy < 1? Let xy be "a". Since both x and y are integers, xy or "a" is also an integer.
Consider (1) alone.
It says xyz > 1or (xy)z >1 or az > 1.
Since both a and z are integers this is possible only if both are positive or both are negative.
Take an example. Let a = 1 and z = 6. Then az = 1*6 = 6 > 1.
Let a = -1 and z = -6. Then az = 6 > 1
So if a >0, we have that a >= 1, because it is an integer and so xy >= 1.
If a < 0, we have that a <= 1, because it is an integer and so xy<=1.
Since nothing definite can be said about xy, (1) is not enough.
Consider (2) alone.
If xy(z^2) > 1, we have that a(z^2) > 1. Since z^2 is positive, being a perfect square, a >= 1 or xy >= 1. So xy is not less than 1.
Hence (2) is enough.

The correct answer is (B).