1. The problem is to find the derivative $\frac{dy}{dx}$ given the implicit equation $$xy + y^2 = 4.$$ 2. We use implicit differentiation, differentiating both sides with respect to $x$. Recall the product rule: $$\frac{d}{dx}(uv) = u'v + uv'$$ and the chain rule for $y^2$: $$\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}.$$ 3. Differentiate the left side: $$\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4).$$ 4. Applying the product rule to $xy$: $$\frac{d}{dx}(xy) = x \frac{dy}{dx} + y.$$ 5. Differentiating $y^2$: $$\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}.$$ 6. The right side derivative is zero since 4 is constant: $$0.$$ 7. Substitute these into the equation: $$x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0.$$ 8. Group terms with $\frac{dy}{dx}$: $$(x + 2y) \frac{dy}{dx} + y = 0.$$ 9. Isolate $\frac{dy}{dx}$: $$(x + 2y) \frac{dy}{dx} = -y.$$ 10. Divide both sides by $(x + 2y)$: $$\frac{dy}{dx} = \frac{-y}{x + 2y}.$$ 11. Show cancellation explicitly: $$\frac{dy}{dx} = \frac{\cancel{-y}}{\cancel{x + 2y}} = \frac{-y}{x + 2y}.$$ 12. The derivative is $$\boxed{\frac{dy}{dx} = \frac{-y}{x + 2y}}.$$ 13. Comparing with the options, the correct answer is a. $-\frac{y}{x+2y}$.