1. **Problem statement:** Find the limit $$\lim_{(x,y)\to(0,0)} \frac{xy^2}{x^2 + y^2}$$ and analyze it along the curve $$y = mx$$ where $$m$$ is a constant slope. 2. **Substitute $$y = mx$$ into the function:** $$f(x,mx) = \frac{x(mx)^2}{x^2 + (mx)^2} = \frac{x m^2 x^2}{x^2 + m^2 x^2} = \frac{m^2 x^3}{x^2(1 + m^2)}$$ 3. **Simplify the expression:** $$f(x,mx) = \frac{m^2 x^3}{x^2 (1+m^2)} = \frac{m^2 x}{1 + m^2}$$ 4. **Evaluate the limit as $$x \to 0$$:** $$\lim_{x \to 0} f(x,mx) = \lim_{x \to 0} \frac{m^2 x}{1 + m^2} = 0$$ 5. **Interpretation:** The limit along any line $$y=mx$$ approaching $$(0,0)$$ is 0. 6. **Check if the limit exists in general:** Consider polar coordinates: $$x = r\cos\theta, y = r\sin\theta$$ 7. Substitute: $$f(r,\theta) = \frac{r\cos\theta (r\sin\theta)^2}{r^2} = \frac{r \cos\theta r^2 \sin^2\theta}{r^2} = r \cos\theta \sin^2\theta$$ 8. **Evaluate:** $$\lim_{r \to 0} r \cos\theta \sin^2\theta = 0$$ for all $$\theta$$. 9. Since the limit is 0 independently of $$\theta$$, the limit exists and is 0. **Final answer:** $$\boxed{0}$$