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