Given \( x = 2\sin 60^\circ \cos 60^\circ \), \( y = \sin^2 30^\circ - \cos^2 30^\circ \), and \( x^2 = ky^2 \).
Calculate \( x \):
\( x = 2\sin 60^\circ \cos 60^\circ = 2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = 2\left(\frac{\sqrt{3}}{4}\right) = \frac{\sqrt{3}}{2} \).
Calculate \( y \):
\( y = \sin^2 30^\circ - \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} - \frac{3}{4} = \frac{-2}{4} = -\frac{1}{2} \).
Substitute \( x \) and \( y \) into \( x^2 = ky^2 \):
\( \left(\frac{\sqrt{3}}{2}\right)^2 = k\left(-\frac{1}{2}\right)^2 \)
\( \frac{3}{4} = k\left(\frac{1}{4}\right) \)
\( 3 = k \)
Therefore, \( k = 3 \).
Final Answer: \( \boxed{3} \)