Answer
$$\frac{{{u^2} - 9}}{{{u^2} + 9}}$$
Work Step by Step
$$\eqalign{
& \cos \left( {2{{\tan }^{ - 1}}\frac{3}{u}} \right) \cr
& {\text{From the triangle we have that}} \cr
& \tan \theta = \frac{3}{u} \cr
& \theta = {\tan ^{ - 1}}\frac{3}{u} \cr
& \cos \left( {2{{\tan }^{ - 1}}\frac{3}{u}} \right) = \cos \left( {2\theta } \right) \cr
& \cos \left( {2{{\tan }^{ - 1}}\frac{3}{u}} \right) = 1 - 2{\sin ^2}\theta \cr
& \cos \left( {2{{\tan }^{ - 1}}\frac{3}{u}} \right) = 1 - 2{\left( {\frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}}} \right)^2} \cr
& \cos \left( {2{{\tan }^{ - 1}}\frac{3}{u}} \right) = 1 - 2{\left( {\frac{{\text{3}}}{{\sqrt {{u^2} + 9} }}} \right)^2} \cr
& \cos \left( {2{{\tan }^{ - 1}}\frac{3}{u}} \right) = 1 - \frac{{18}}{{{u^2} + 9}} \cr
& \cos \left( {2{{\tan }^{ - 1}}\frac{3}{u}} \right) = \frac{{{u^2} - 9}}{{{u^2} + 9}} \cr} $$
