Let #(1+3x)=t, and, ln(1+3x)=lnt=u#, so that,
#y=ln(ln(1+3x)=ln(lnt)=lnu#
Thus, #y# is a function of #u#, #u# of #t#, and #t# of #x#.
Accordingly, by The Chain Rule,
#dy/dx#=#(dy)/(du)##(du)/(dt)##(dt)/(dx)#.................................#(star)#
#y=lnu rArr dy/(du)=1/u..........................(1)#
#u=lnt rArr (du)/dt=1/t..............................(2)#
#t=1+3x rArr (dt)/dx=3..............................(3)#
Therefore, by #(1),(2),(3), and, (star)#, we get,
#dy/dx=1/u*1/t*3=3/(tu)#
#:. dy/dx=3/((1+3x)ln(1+3x))#.
Enjoy Maths.!