This is the solution that I found. I do not find this explanation satisfactory.

Can any of the experts please explain the solution better? Thanks in advance

Let's start by simplifying the original by multiplying both sides by y² (we can safely do so because we know that y² is positive):
Is \(x^2y^3\) - \(y^3\) > 1?
(1) x < y
If x and y are big positive numbers, \(y^3\) will dominate x²y² and the left side will be negative; if x and y are big negative numbers, x²y² and (-\(y^3\)) will both be positive and the left side will be a big positive number: insufficient.
(2) x² < y²
Tells us that |x| < |y|, but nothing about the signs of x and y. If we pick really big values with different signs we can generate both "yes" and "no" answers.
Together: combining doesn't help us at all, since we have the same generalizations from both: choose E.
_________________

Please +1 KUDO if my post helps. Thank you.

\(x^2\) - y > 1/\(y^2\)

\(x^2\) - y - 1/\(y^2\) > 0

Lets try 2 first
(2) x² < y² or y^2 > x^2,
9 > 4, y can be -3 or 3 and x can be -2 or 2
1/4 > 1/16, y can be 1/2 or -1/2 and x can be 1/4 or -1/4

4 +3 - 1/9 > 0, This answers a yes to the question
1/16 - 1/2 - 1/4 < 0, This answers a no to the question

from 1)
x < y or y>x, here y & x can be again a fraction, integer or a number
1/2 > 1/4 or 3 > 2
1/16 - 1/2 - 1/4 < 0, This answers a no to the question
4 -3 - 1/9 > 0, This answers a yes to the question

The same test cases can be used again to show that after combining the 2, we wont get an unique answer

E
_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Take (x,y) = (0,1) & (0,-1).

We get E.
_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

Here's how I solved, feel free to critique!

Is \(x^{2}\) - y > \(\frac{1}{y^{2}}\)? If \(y\neq{0}\).

I simplified the expression as follows:

Because \(y^{2}\) is always positive I multiplied both sides by \(y^{2}\) to get \(x^{2}y^{2}\) - \(y^{3}\) > 1.

From here I factored out \(y^{2}\) to get \(y^{2}\)(\(x^{2}\) - y) > 1.

The question is now:

Is \(y^{2}\)(\(x^{2}\) - y) > 1? If Y \(x\neq{0}\).

Statement 1: x < y

I picked x as -1 and y as 1. Plugging these values into the question the result is 0 > 1.

I then picked x as -3 and y as -2. Plugging these values into the question the result is 44 > 1.

Clearly insufficient.

Statement 2: \(x^{3}\) < \(y^{3}\)

I picked x as -1 and y as 1. Since they are raised to an odd power the signs remain the same for x and y (1 and -1, respectively).
Plugging these values into the question the result is 0 > 1.

I then picked x as -3 and y as -2. Since they are raised to an odd power the signs remain the same for x and y (-3 and -2, respectively).
Plugging these values into the question the result is 44 > 1.

Clearly insufficient.

Statement 1&2

Clearly insufficient.

Therefore, the answer is E.
_________________

...once in a while you get shown the light, in the strangest of places if you look at it right...