\(x^2 * y^2 = 18 – 3xy,\)
Lets rewrite the equation as \((xy)^2 + 3xy - 18= 0\)
Now lets take xy as T we get
\(T^2 + 3T - 18= 0\)
\(T^2 + 6T - 3T - 18= 0\)
\(T(T + 6) - 3(T + 6)= 0\)
\((T + 6)(T - 3)= 0\)
=> T = 3, -6
=> xy = 3 or xy =-6
=> x = \(\frac{3}{y}\)
Squaring both sides we get
\(x^2\) = \(\frac{9}{y^2}\)

So, answer will be D
Hope it helps!