How do you differentiate #y = (ln(x^2))^(2x+3)#?

Note firstly that #y# can be expressed more simply.
#y=(ln(x^2))^(2x+3)#
#y=(2lnx)^(2x+3)#

Take the variable out of the exponent by taking logarithms:
#lny=(2x+3)ln(2lnx)#

Differentiate with the product rule (RHS) and the chain rule (LHS):
#1/ydy/dx=2ln(2lnx)+(2x+3)d/dx[ln(2lnx)]#

Differentiate the double log by the chain rule:
#d/dx[ln(2lnx)]=1/(2lnx)*d/dx[2lnx]=1/(2lnx)*2/x=1/(xlnx)#

So
#1/ydy/dx=2ln(2lnx)+(2x+3)/(xlnx)#

Thus
#dy/dx=(2lnx)^(2x+3)[2ln(2lnx)+(2x+3)/(xlnx)]#
or, expressed in the fashion of the question
#dy/dx=(ln(x^2))^(2x+3)[2lnln(x^2)+(2x+3)/(xlnx)]#

We seek the derivative of:

# y = (ln(x^2))^(2x+3) #

We can take Natural Logarithms of both sides:

# ln y = ln {(ln(x^2))^(2x+3)} #

And using the logarithm properties, #lna^b=blna# and #lnab=lna+lnb#, this becomes:

# ln y = (2x+3)ln {ln(x^2)} #

# \ \ \ \ \ \ = (2x+3)ln {2lnx} #

# \ \ \ \ \ \ = (2x+3){ln 2 + ln(lnx)} #

Then by applying the product rule, and differentiating implicitly, we have:

# 1/y \ dy/dx = (2x+3)d/dx{ln 2 + ln(lnx)} + d/dx{(2x+3)} \ {ln 2 + ln(lnx)} #

Applying the chain rule we get:

# 1/y \ dy/dx = (2x+3){1/(lnx)d/dx(lnx)} + 2 {ln 2 + ln(lnx)} #

# \ \ \ \ \ \ \ \ \ \ = (2x+3)/(xlnx) + 2 ln 2 + 2ln(lnx) #

Leading to:

# dy/dx = y{(2x+3)/(xlnx) + 2 ln 2 + 2ln(lnx)} #

# \ \ \ \ \ = (ln(x^2))^(2x+3){(2x+3)/(xlnx) + 2 ln 2 + 2ln(lnx)} #