We know that \(x^3>y^2\) and \(y>0\) and we need to tell which all options are true

I. \(x>y\)
II. \(x>0\)
III. \(\frac{x}{y}<0\)

Now we know that \(x^3>y^2\) and we also know that square of any number can never be negative so \(y^2\) >= 0 ( it is actually >0 as y > 0)
=> \(x^3>y^2\) => \(x^3>0\). Take cube root on both the sides we get
x > 0

So, we know that x > 0 and y > 0
=> \(\frac{x}{y}\) > 0(as \(\frac{Positive}{Positive}\) > 0)

Let's take the option choices now

I. \(x>y\)
Now, we know that both x and y are positive and \(x^3>y^2\), but this doesn't mean that x is ALWAYS > y
Case 1. x = 3 , y = 4
\(x^3\) = \(3^3\) = 27 > \(y^2\) which is 16
So, \(x^3>y^2\) but x < y

Case 2. x = 10, y =2
Clearly \(x^3>y^2\) is true
So, \(x^3>y^2\) but x > y
So, x > y might not be true always
So, FALSE

II. \(x>0\)
We proved above that x > 0
So, TRUE

III. \(\frac{x}{y}<0\)
We proved above that \(\frac{x}{y}>0\)
So, FALSE

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Inequalities