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]