How do you find the derivative of #(ln tan(x))^2#?

How? Use the chain rule -- twice.
And simplify the answer. (Often simplifying is the tricky part.)

The derivative of some stuff squared is two times the stuff, times the derivative of that stuff.
Formally : the derivative of #[g(x)]^2# is #2[g(x)]*g'(x)#

In this case the "inside stuff", the #g(x)# is itself, the #ln# of some stuff. So to find the derivative, we'll need the chain rule again.
The derivative of the #la# of some stuff is #1# over the stuff, times the derivative of that stuff.
Formally : the derivative of #ln[f(x)]# is #1/f(x)*f'(x)#

Finally, we need to know the derivative of #tanx#.

The derivative of #(ln tan(x))^2# is

#2(ln tan(x))*1/tan(x)*sec^2(x)#

Now, we've finished the calculus, but we can re-write our answer in simpler form.
#1/tan(x) = cot(x)#
and #cot(x) = cos(x)/sin(x)#.

The second equality wouldn't be helpful, except that #sec(x)=1/cos(x)#.

Therefore we can re-write the derivative using:

#1/tan(x)*sec^2(x) = cot(x)sec^2(x)=cos(x)/sin(x)*1/cos^2(x)=1/sin(x)*1/cos(x)#

Which is surely more simply written as #csc(x)sec(x)#

So
The derivative of #(ln tan(x))^2# is

#2(ln tan(x))csc(x)sec(x)#

(If there's any advantage to it, we could also re-write #ln tan(x)# as #ln(sinx)-ln(cosx)#.)