Step 1: Rewrite:
Let $2a = \theta$. Then $\cos\theta = \frac{x^2-y^2}{x^2+y^2}$, which gives $\tan^2\frac{\theta}{2} = \frac{1-\cos\theta}{1+\cos\theta} = \frac{2y^2}{2x^2}$.
Step 2: Slope:
So $y = x\tan a$, a straight line through the origin, and its slope is $\tan a$. Therefore $y\,y' = (x\tan a)(\tan a) = x\tan^2 a$.
Step 3: Evaluate:
$\tan\frac{2\pi}{3} = -\sqrt3$, and squaring gives 3.
Final Answer:
The answer is 3, option (C).
\[ \boxed{3} \]