We have: #f(x) = x^(3) - 3 x^(2)#

First, let's find the #y#-coordinate corresponding to the given value of #x#:

#Rightarrow f(4) = (4)^(3) - 3 cdot (4)^(2)#

#Rightarrow f(4) = 64 - 48#

#therefore f(4) = 16#

So we must find the equation of the normal line at the point #(4, 16)#.

Then, let's differentiate #f(x)#:

#Rightarrow f'(x) = 3 x^(2) - 6 x#

For #x = 4#, the slope of the line is:

#Rightarrow f'(4) = 3 cdot (4)^(2) - 6 cdot (4)#

#Rightarrow f'(4) = 48 - 24#

#therefore f'(4) = 24#

The gradient of the normal line is the reciprocal of this value:

#Rightarrow "Gradient of normal line" = (f'(4))^(- 1)#

#therefore "Gradient of normal line" = frac(1)(24)#

Now, let's express the equation of the normal line in point-slope form:

#Rightarrow y - y_(1) = m (x - x_(1))#

#Rightarrow y - 16 = frac(1)(24) (x - 4)#

#Rightarrow y - 4 = frac(1)(24) x - frac(1)(6)#

#therefore y = frac(1)(24) x - frac(23)(6)#

Therefore, the equation of the normal line is #y = frac(1)(24) x - frac(23)(6)#.