1. The problem asks to differentiate the function $$y=\frac{x^2}{x^2+1}$$ with respect to $$x$$. 2. We use the quotient rule for differentiation: if $$y=\frac{u}{v}$$, then $$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$. 3. Here, $$u=x^2$$ and $$v=x^2+1$$. 4. Differentiate $$u$$ and $$v$$: $$\frac{du}{dx}=2x$$ $$\frac{dv}{dx}=2x$$ 5. Apply the quotient rule: $$\frac{dy}{dx}=\frac{(x^2+1)(2x)-x^2(2x)}{(x^2+1)^2}$$ 6. Simplify the numerator: $$(x^2+1)(2x)-x^2(2x)=2x^3+2x-2x^3=2x$$ 7. So, $$\frac{dy}{dx}=\frac{2x}{(x^2+1)^2}$$ Final answer: $$\boxed{\frac{dy}{dx}=\frac{2x}{(x^2+1)^2}}$$