#### Answer

$a.$ The value of the derivative is $f'(3)=6$.
$b.$ The equation is $y=6x-9$.

#### Work Step by Step

$a.$ By definition of a derivative at point $a=3$ we have
$$f'(3)=\lim_{h\to0}\frac{f(3+h)-f(3)}{h}=\lim_{h\to0}\frac{(3+h)^2-3^2}{h}=\lim_{h\to0}\frac{9+h^2+6h-9}{h}=\lim_{h\to0}\frac{h^2+6h}{h}=\lim_{h\to0}(h+6)=0+6=6.$$
$b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=3$, $f(a)=3^2=9$ and $f'(a)=6$ we get
$$y-9=6(x-3)\Rightarrow y-9=6x-18$$ which gives $$y=6x-9.$$