Lets use some logic:
From the stem:
Y is some positive fraction
X is either some positive integer or some negative integer

A) No - if x<0, it is not >1
B) If x>0, yes but if x<0, no (consider: y=1/2 and x=-2 x=2)
C) If x<0, this will be something negative
D) This is only true for some values of x - for example, if x=5, and y=1/2 but if x=1, y=1/4 it is no.
E) Last choice and it must be correct - x^2 is always positive, dividing by a fraction is the same as multiplying by its reciprocal, which in the case of y will be some positive integer.

0<y<1, so y is a positive fraction
x is a non-zero integer, so, x could be either positive or negative.

Just plugging in any negative ineteger values for x, say -2, we can eliminate A, B, and C.

For D and E, x^2 is always positive regardless of the value of x.

For D, if x=2 and y = 1/2, we get (2)^2*(1/2)=2, which is >1
However, if x=1, then (1)^2*1/2=1/2, which is < 1.

So, only option left is E.

The logic behind E is if we divide an integer by a positive fraction, it's always greater than 1, because big/small=bigger

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The first thing to note is that \(y\) is a fraction. If an integer is divided by a fraction we can always be sure that it will be greater than \(1\).

\(\frac{x}{y}\) and \(\frac{x^2}{y}\), Choices B and E are possible answers.

But if \(x\) is a non-zero integer, it can be positive or negative, and dividing a negative integer by a fraction will give us a bigger negative integer which will be less than \(1\). So we need to ensure that \(y\) divides a positive integer.

That is possible only with Choice E, \(\frac{x^2}{y}\), as \(x^2\) will ensure us that the number being divided is positive.