Solution:

Fixing $y,$ solve $\sin^2 x - \sin x \sin y + \sin^2 y = \dfrac34$ as a quadratic in $\sin x:$ $$\sin x = \frac{1}{2}\sin y \pm \frac{\sqrt3}{2}\cos y = \sin\!\left(y \pm \frac{\pi}{3}\right).$$

Within $S,$ $\sin x = \sin\!\left(y - \tfrac{\pi}{3}\right)$ gives the line $x = y - \tfrac{\pi}{3},$ while $\sin x = \sin\!\left(y + \tfrac{\pi}{3}\right)$ gives $x = y + \tfrac{\pi}{3}$ for $y \le \tfrac{\pi}{6}$ and $x = -y + \tfrac{2\pi}{3}$ for $y \ge \tfrac{\pi}{6}.$

These lines split $S$ into regions; testing the corners shows the inequality holds only in the middle band. Its area is $$\left(\frac{\pi}{2}\right)^2 - \frac{1}{2}\left(\frac{\pi}{3}\right)^2 - 2 \cdot \frac{1}{2} \left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{6}.$$

Thus, the correct answer is C.