Here since no value are give let us take

both x and y be positive integer such that x= 3 and y =2

From statement 1: $x^2 \div (y+\frac{1}{y})$ = $\frac{18}{5}$ = 3.6 (putting the value of x and y in the equation)

From Statement 2 : $y^2 \div (x+\frac{1}{x})$ = $\frac{12}{10}$= 1.2 (putting the value of x and y in the equation)

From above we get statement 1 > Statement 2

Now if we consider negative integer such that x= -2 and y = -3 (since x>y)

From statement 1: $x^2 \div (y+\frac{1}{y})$ = $\frac{- 12}{10}$= -1.2 (putting the value of x and y in the equation)

From Statement 2 : $y^2 \div (x+\frac{1}{x})$ = $\frac{- 18}{5}$ = -3.6(putting the value of x and y in the equation)

From above we have Statement 1 > Statement 2

But we have to look into every possibilities

Now if we consider negative integer such that x= 2 and y = -2 (since x>y)

From statement 1: $x^2 \div (y+\frac{1}{y})$ = $\frac{- 4}{5}$= -0.8 (putting the value of x and y in the equation)

From Statement 2 : $y^2 \div (x+\frac{1}{x})$ = $\frac{4}{5}$ = 0.8(putting the value of x and y in the equation)

From above we have Statement 2 > Statement 1.

Hence the option is D