Note that #ln(x+3)# is not defined if #x+3<0# i.e. #x<-3#.

Further if #x=-2#, #g(-2)=ln(-2+3)=ln1=0# and therefore in the interval #(-2,oo)# #ln(x+3)# is above #x#-axis and in interval #(-3,-2)#, it is below #x#-axis.

Also #g'(x)=1/(x+3)# and hence in the interval #(-3,oo)#, #g'(x)# is always positive and hence #g(x)# is a continuously increasing function and its graph looks like

graph{ln(x+3) [-10, 10, -5, 5]}

Hence, for #y=|ln(x+3)|#,

While in the interval #(-2,oo)#, it is exactly same as for #ln(x+3)#, in interval #(-3,-2)# it will be positive too and a reflection of curve above. Obviously, we have a discontinuity at #x=-2#

graph{|ln(x+3)| [-10, 10, -5, 5]}