Rewrite 2tan 3x in terms of tan x.

We shall use the identity

tan (x+y) = (tanx + tany) / (1 - tanx tany)

Therefore

tan (3x) = tan (2x+x) = [tan (2x) + tan (x) ]/[1 - tan (2x) tan (x) ]

tan (2x) = (tanx + tanx) / (1 - tan^2x)

That is

tan (3x) = [ (2tanx / (1-tan^2x} + tanx]/[1 - tanx (2tanx) / (1-tan^2x) ]

= [2tanx + tanx (1-tan^2x) ][ (1-tanx) (1-tan^2x) + 2 tan^2x]

= [2tanx + tanx - tan^3x]/[1-tan^2x - tanx + tan^3x + 2tan^2x]

= [3tanx - tan^3x]/[1 - tanx + tan^2x + tan^3x]

2 tan3x = [6tanx - 2tan^3x]/[1 - tanx + tan^2x + tan^3x]

The answer is 2 (3tan (x) - tan^3 (x)) / (1-3tan^2 (x)) but it is a very long and tedious calculation using alot of trig identities.