Answer
$a)$ $f(x+h)=\frac{1}{(x+h)^2}$
$b)$ $f(x+h)-f(x)=\frac{-2xh-h^2}{(x+h)^2x^2}$
$c)$ $\frac{f(x+h)-f(x)}{h}=\dfrac{-2x-h}{(x+h)^2x^2}$
Work Step by Step
$a)$
$$f(x)=\frac{1}{x^2}$$
Replace $x$ with $x+h$:
$$f(x+h)=\frac{1}{(x+h)^2}$$
$b)$
$$f(x+h)-f(x)$$
Substitute corresponding expressions:
$$\frac{1}{(x+h)^2}-\frac{1}{x^2}$$
$$\frac{x^2-(x+h)^2}{(x+h)^2x^2}=\frac{x^2-x^2-2xh-h^2}{(x+h)^2x^2}=\frac{-2xh-h^2}{(x+h)^2x^2}$$
$c)$
$$\frac{f(x+h)-f(x)}{h}$$
Substitute corresponding expression for numerator:
$$\frac{\dfrac{-2xh-h^2}{(x+h)^2x^2}}{h}=\frac{\dfrac{h(-2x-h)}{(x+h)^2x^2}}{h}=\dfrac{-2x-h}{(x+h)^2x^2}$$
