Let, #f(x)=ln(x^8)/x^2=(8lnx)/x^2.....[because, ln a^m=mlna],#
#:. f(x)=8*x^-2*lnx.#
By the Product Rule, then, we get,
# f'(x)=8{x^-2d/dx(lnx)+(lnx)d/dx(x^-2)},#
#=8{x^-2*1/x+(-2*x^(-2-1))lnx},#
#=8{x^-1-2x^-3*lnx},#
#=8{1/x-(2lnx)/x^3},#
#rArr f'(x)=8/x^3(x^2-2lnx).#
Knowing that, #f''(x)={f'(x)}',# we have,
#f''(x)={8(x^-1-2x^-3*lnx)}',#
#=8(x^-1-2x^-3*lnx)',#
#=8(x^-1)'-8*2(x^-3*lnx)',#
#=8(-1*x^(-1-1))-16{x^-3d/dx(lnx)+(lnx)d/dx(x^-3)},#
#=-8x^-2-16{x^-3*1/x+(-3*x^(-3-1))lnx},#
#=-8/x^2-16(1/x^2-(3lnx)/x^4),#
#=-8/x^2-16/x^2+(48lnx)/x^4,#
#=-24/x^2+(48lnx)/x^4.#
#rArr f''(x)=24/x^4(2lnx-x^2).#
Enjoy Maths.!