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)