Answer
The difference quotient of $f\left( x \right)=\frac{1}{{{x}^{2}}}$is$-\dfrac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}$.
Work Step by Step
The difference quotient of $f$ at $x$ is given by$\frac{f\left( x+h \right)-f\left( x \right)}{h},\,\,h\ne 0$.
$f\left( x+h \right)=\frac{1}{{{\left( x+h \right)}^{2}}}$ and$f\left( x \right)=\frac{1}{{{x}^{2}}}$,
Plug in these values,
$\frac{f\left( x+h \right)-f\left( x \right)}{h}=\frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}$
Subtract,
$=\frac{\frac{{{x}^{2}}-{{\left( x+h \right)}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}}}{h}$
$=\frac{\frac{{{x}^{2}}-\left( {{x}^{2}}+2xh+{{h}^{2}} \right)}{{{x}^{2}}{{\left( x+h \right)}^{2}}}}{h}$
Divide and distribute,
$=\frac{{{x}^{2}}-{{x}^{2}}-2xh-{{h}^{2}}}{h{{x}^{2}}{{\left( x+h \right)}^{2}}}$
Simplify,
$=\frac{-2xh-{{h}^{2}}}{h{{x}^{2}}{{\left( x+h \right)}^{2}}}$
Factor out $h$ from the numerator,
$=\frac{h\left( -2x-h \right)}{h{{x}^{2}}{{\left( x+h \right)}^{2}}}$
Dividing out the factor$h$,
$=-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}$
Hence, the difference quotient of $f\left( x \right)=\frac{1}{{{x}^{2}}}$ is$-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}$.
