Answer
$-\displaystyle \frac{2x+h}{x^{2}(x+h)^{2}}$
Work Step by Step
$f(x)=\displaystyle \frac{1}{x^{2}}$
$\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}$
... Write the numerator as one fraction, LCD=$x^{2}(x+h)^{2}$
$=\displaystyle \frac{1}{h}\cdot\frac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}}$
$=\displaystyle \frac{1}{h}\cdot\frac{x^{2}-(x^{2}+2xh+h^{2})}{x^{2}(x+h)^{2}}$
$=\displaystyle \frac{1}{h}\cdot\frac{-2xh-h^{2}}{x^{2}(x+h)^{2}}$ ... factor out $-h$ in the numerator
$=\displaystyle \frac{1}{h}\cdot\frac{-h(2x+h)}{x^{2}(x+h)^{2}}$ ... h cancels
$=-\displaystyle \frac{2x+h}{x^{2}(x+h)^{2}}$
