#f(x)=ln(3x)/x^2-x^2#
equation of tangent is #y=mx+c#
m is the slope and y intercept c
intercept point is (3,f(3)
#f'(x)=d/dx(ln(3x)/x^2-x^2)#
#f'(x)=d/dx(ln(3x)/x^2)-d/dx(x^2)#
#f'(x)=(x^2*d/dx(ln(3x))-ln(3x)d/dx(x^2))/x^4-d/dx(x^2)#
#f'(x)=((x^2*3/(3x))-ln(3x)(2x))/x^4-2x#
#f'(x)=(x-2xln(3x))/x^4-2x=(1-2ln(3x))/x^3-2x#
slope at x=3
#f'(3)=(1-2ln(3(3)))/(3)^3-2(3)#
#f'(3)=(1-2ln(9))/27-6#
#f(3)=ln(9)/9-9#
intercept point is #(3,ln(9)/9-9)#
equation of line at point #(3,ln(9)/9-9)#
#y=((1-2ln(9))/27-6)x+c#
to find value of c put #(3,ln(9)/9-9)# in above equation
#ln(9)/9-9=((1-2ln(9))/27-6)(3)+c#
#c=ln(9)/9-9-((1-2ln(9))/27-6)(3)#
#c=ln(9)/9-9-3/27+6/27ln(9)+18#
#c=ln(9)/9-9-1/9+2/9ln(9)+18#
#c=ln(9)/9+80/9+2/9ln(9)#
Hence Equation of slope is
#y=((1-2ln(9))/27-6)x+ln(9)/9+80/9+2/9ln(9)#
