#### Answer

$f'(a) = -\dfrac{2}{a^3}$

#### Work Step by Step

Given $f(x)=x^{-2}$
$f'(a) = \lim\limits_{x \to a} \dfrac{f(x) - f(a)}{x - a} = \lim\limits_{x \to a} \dfrac{x^{-2} - a^{-2}}{x - a} = \lim\limits_{x \to a} \dfrac{\frac{1}{x^2} - \frac{1}{a^2}}{x - a} = \lim\limits_{x \to a} \dfrac{\frac{a^2 - x^2}{a^2x^2}}{x - a} = \lim\limits_{x \to a} \dfrac{a^2-x^2}{a^2x^2(x - a)} = \lim\limits_{x \to a} \dfrac{-(x^2-a^2)}{a^2x^2(x - a)} = \lim\limits_{x \to a} \dfrac{-(x+a)(x-a)}{a^2x^2(x - a)} = \lim\limits_{x \to a} \dfrac{-(x+a)}{a^2x^2} = \dfrac{-(a+a)}{a^2a^2} = \dfrac{-(2a)}{a^{2+2}} = -\dfrac{2a}{a^4} = -\dfrac{2}{a^3} \longrightarrow f'(a) = -\dfrac{2}{a^3}$