Answer
$$f'\left( x \right) = - \frac{{3x}}{{9 + {x^2}}} + {\cot ^{ - 1}}\left( {\frac{x}{3}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x{\cot ^{ - 1}}\left( {x/3} \right) \cr
& {\text{find the derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {x{{\cot }^{ - 1}}\left( {x/3} \right)} \right] \cr
& {\text{use product rule}} \cr
& f'\left( x \right) = x\frac{d}{{dx}}\left[ {{{\cot }^{ - 1}}\left( {\frac{x}{3}} \right)} \right] + {\cot ^{ - 1}}\left( {\frac{x}{3}} \right)\frac{d}{{dx}}\left[ x \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{{\cot }^{ - 1}}u} \right] = - \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}.{\text{ let }}u = \frac{x}{3} \cr
& f'\left( x \right) = x\left( { - \frac{1}{{1 + {{\left( {x/3} \right)}^2}}}} \right)\frac{d}{{dx}}\left[ {\frac{x}{3}} \right] + {\cot ^{ - 1}}\left( {\frac{x}{3}} \right)\left( 1 \right) \cr
& f'\left( x \right) = - \frac{x}{{1 + {{\left( {x/3} \right)}^2}}}\left( {\frac{1}{3}} \right) + {\cot ^{ - 1}}\left( {\frac{x}{3}} \right) \cr
& {\text{simplifying}} \cr
& f'\left( x \right) = - \frac{x}{{1 + {x^2}/9}}\left( {\frac{1}{3}} \right) + {\cot ^{ - 1}}\left( {\frac{x}{3}} \right) \cr
& f'\left( x \right) = - \frac{{9x}}{{9 + {x^2}}}\left( {\frac{1}{3}} \right) + {\cot ^{ - 1}}\left( {\frac{x}{3}} \right) \cr
& f'\left( x \right) = - \frac{{3x}}{{9 + {x^2}}} + {\cot ^{ - 1}}\left( {\frac{x}{3}} \right) \cr} $$
