(a) \[\frac{dy}{dx}=\arctan(3x)+\frac{3x}{1+9x^2}.\] At \(x=1/3\), \(\boxed{dy/dx=\pi/4+1/2}\).

(b) Integrating by parts with \(u=\arctan(3x)\), \(dv=x\,dx\),

\[\begin{aligned}\int x\arctan(3x)\,dx&=\frac{x^2}2\arctan(3x)-\frac12\int\frac{3x^2}{1+9x^2}\,dx\\&=\frac{x^2}2\arctan(3x)-\frac x6+\frac1{18}\arctan(3x)+C,\end{aligned}\]

where \(3x^2/(1+9x^2)=1/3-1/[3(1+9x^2)]\). Applying the limits, \[\boxed{\int_0^{1/3}y\,dx=\frac\pi{72}-\frac1{18}+\frac\pi{72}=\frac\pi{36}-\frac1{18}}.\]