Well, first let's make it look nicer...
#y = 2^(3^(x^2))#
Let #color(darkred)(u = 3^(x^2))#
#=> y = 2^u#
Using logarithmic simplification:
#ln_2(y) = u -> (lny)/(ln2) = u -> lny = u ln2#
And now with some manipulation:
#y = e^(u ln2) = e^(ln2^u) = 2^u#
#color(green)((dy)/(du) = e^(u ln2)*ln2 = 2^u ln2)#
Similarly:
Let #color(darkred)(w = x^2)#
#=> u = 3^w#
#ln_3(u) = w -> (lnu)/(ln3) = w -> lnu = wln3#
#u = e^(wln3) = e^(ln3^w) = 3^w#
#color(green)((du)/(dw) = e^(wln3)*ln3 = 3^w ln3)#
Lastly, #color(green)((dw)/(dx) = 2x)#.
Thus, taking the whole derivative:
#d/(dx)[y = 2^(u(w(x)))] = d/(dx)[y = 2^(u(x^2))] = d/(dx)[y = 2^(3^(x^2))] = ((dy)/(du)) ((du)/(dw)) ((dw)/(dx))#
#= (2^u ln2)(3^w ln3)(2x)#
#= (2^(3^(x^2)) ln2)(3^(x^2) ln3)(2x)#
#= 2xln2ln3(2^(3^(x^2))3^(x^2))#
#= color(blue)(xln2ln3(2^(3^(x^2) + 1)3^(x^2)))#
