I will assume that you know, in general how to find inflection points.
#f(x)=(1+lnx)^3#
#f'(x) = 3/x (1+ ln x)^2#
#f''(x)=-3/x^2(1+ln x)^2 color(red)(+) 3/x*2(1+ ln x) * 1/x#
#=3/x^2(1+ ln x)[-(1+ln x) color(red)(+)2]#
#=3/x^2(1+ ln x)(1-ln x)=3/x^2(1-ln^2 x)#
#f''(x)# does not exist for #x <= 0# and neither does #f(x)#, so the question of concavity there cannot arise.
Solve: #f''(x)=0#
#3/x^2(1-ln^2 x)=0#
at #ln x = +-1#, so, #x=e# or #x= 1/e#
Because #3/x^2 > 0# for all #x#,
the sign of #f''# depends only on the sign of #1-ln^2x#
On #(0, 1/e)#, we have #ln x < -1# so #ln^2x > 1# and #1-ln^2x <0# #f''(x) <0#
On #(1/e, e)#, we have #-1< ln x < 1# so #0< ln^2x < 1# and #1-ln^2x > 0# so #f''(x) > 0#
The concavity changes at #(1/e, f(1/e)) = (1/e, 0)#
On #(e, oo)#, we have #ln x > 1# so #ln^2x > 1# and #1-ln^2x <0# #f''(x) <0#
The concavity changes at #(e, f(e)) = (e, 8)#
The inflection points are #(1/e, 0)# and #(e, 8)#
