We can rewrite the equation as follows (and consequently avoid needing to use the product rule):
# f(x) = ln(x+xe^(3x)) #
# :. f(x) = ln(x(1+e^(3x))) #
# :. f(x) = ln(x) + ln(1+e^(3x)) #
We can differentiate (using the chain rule):
# :. f'(x) = 1/x + 1/(1+e^(3x))*(3e^(3x)) #
# :. f'(x) = 1/x + (3e^(3x))/(1+e^(3x)) #
# :. f'(x) = ( (1+e^(3x)) + (3xe^(3x)) ) / (x(1+e^(3x)))#
# :. f'(x) = ( 3xe^(3x) + e^(3x) + 1) / (x + xe^(3x))#
When #x=2=>f(2)=ln(2+2e^6) #
And, #f'(2)=(6e^6 + e^6 +1)/(2+2e^6) = (1+7e^6)/(2+2e^6)#
This is the gradient of the tangent when #x=2#. As the normal and tangent are perpendicular the product of their gradients is #-1#, so the normal passes through #(2,ln(2+2e^6))# and it has gradient #m=-(2+2e^6)/(1+7e^6)#
So using #y-y_1=m(x-x_1)# the required equation is;
# y - ln(2+2e^6) = -(2+2e^6)/(1+7e^6) (x - 2) #
