Derivative of a product involving arctan(x)

Apply the product rule: $\dfrac{d}{dx}[f\cdot g] = f'g + fg'$, with $f(x) = x^2$ so $f'(x) = 2x$, and $g(x) = \arctan(x)$ so $g'(x) = \dfrac{1}{1+x^2}$ by the standard inverse-trig derivative formula.

Substituting into the product rule gives $\dfrac{dy}{dx} = 2x\arctan(x) + x^2\cdot \dfrac{1}{1+x^2}$. This can be left as is or simplified slightly, giving the final answer $\dfrac{dy}{dx} = 2x\arctan(x) + \dfrac{x^2}{1+x^2}$.