Answer
The equation of the normal line is
$$y_n=-\frac{1}{3}x+1.$$
Work Step by Step
Step 1. Find the slope of the tangent at the point $P(3,0)$. By definition of the tangent slope at $P(a,f(a))$ with $a=3$ and $f(a)=0$ we have
$$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to0}\frac{(3+h)^2-3(3+h)-0}{h}=\lim_{h\to0}\frac{9+h^2+6h-9-3h}{h}=\lim_{h\to0}\frac{h^2+3h}{h}=\lim_{h\to0}(h+3)=0+3=3.$$
Step 2. Find the equation of the tangent line. The tangent line at $(a,f(a))$ is given by $y-f(a)=m(x-a)$. Using $a=3$, $f(a)=0$ and $m=3$ we get
$$y-0=3(x-3)\Rightarrow y=3x-9.$$
Step 3. At the point $P(a,f(a))$ when the tangent slope is $m$ the equation of the normal line is $y_n=cx+d$ where
$$m\cdot c=-1,\quad c\cdot a+d=f(a).$$
Using $a=3$, $f(a)=0$, $m=3$ we have
$$3c=-1,\quad 3c+d=0.$$
From the first equation we have $c=-\frac{1}{3}$. Putting this into the second equation we have
$$-\frac{1}{3}\cdot 3+d=0\Rightarrow -1+d=0\Rightarrow d=1.$$
Finally the equation of the normal is
$$y_n=-\frac{1}{3}x+1.$$
