A calculus problem by Zulqarnain Ansari

Let $t = tan\frac{\theta}{2}$ . Then $x = cos\theta, y = sin\theta$ . Then $\frac{dy}{dx} = -cot\theta$ and therefore the given expression is equal to 0.

Using the identity $\ (2xy)^2 +(x^2 -y^2)^2 = (x^2 +y^2)^2$ leads to $\ (1-t)^2 +(2t)^2 = (1+t^2)^2$ $\therefore\ x^2 +y^2 = 1$ Now differentiating implicitly with x gives: $\ 2x +2y.\frac{dy}{dx} = 0$ $\therefore\ y.\frac{dy}{dx} +x = \boxed{0}$