Answer
$f'(x)=\frac{-3}{x^{2}}$
$ f'(-2) = \frac{-3}{4}$
$ f'(0)=0$
$f'(3)=\frac{-1}{3}$
Work Step by Step
$f(x+h)=\frac{3}{x+h}$
$f(x+h) - f(x) =\frac{3}{x+h} - \frac{3}{x}=\frac{3x-3x -3h}{x(x+h)}=\frac{-3h}{x(x+h)}$
$\frac{f(x+h) - f(x)}{h}=\frac{\frac{-3h}{x(x+h)}}{h}=\frac{-3}{x(x+h)}$
$f'(x) = \lim\limits_{h \to 0} \frac{-3}{x(x+h)} =\frac{-3}{x(x+0)}=\frac{-3}{x^{2}}$
$ f'(-2) = \frac{-3}{4}$
$ f'(0)=0$
$f'(3)=\frac{-1}{3}$