I assume you don't know about derivatives yet. The reason I bring that up is because the given limit is equivalent to the derivative of a certain function and you might recognize it later on.
To compute the limit, in the numerator we can combine the fractions into one:
[tex]\dfrac1{(x+h)^2}-\dfrac1{x^2}=\dfrac{x^2-(x+h)^2}{x^2(x+h)^2}=\dfrac{x^2-(x^2+2xh+h^2)}{x^2(x+h)^2}=-\dfrac{(2x+h)h}{x^2(x+h)^2}[/tex]
We're taking the limit as [tex]h\to0[/tex], which means we're considering [tex]h[/tex] near 0, but not [tex]h=0[/tex], so that we can simplify:
[tex]\dfrac{-\frac{(2x+h)h}{x^2(x+h)^2}}h=-\dfrac{2x+h}{x^2(x+h)^2}[/tex]
Then as [tex]h\to0[/tex], we have [tex]2x+h\to2x[/tex] and [tex]x+h\to x[/tex], so that
[tex]\displaystyle\lim_{h\to0}-\frac{2x+h}{x^2(x+h)^2}=-\dfrac{2x}{x^4}=-\dfrac2{x^3}[/tex]
provided that [tex]x\neq0[/tex].
(In fact, the limit is equivalent to the derivative of the function [tex]f(x)=\dfrac1{x^2}[/tex], whose derivative is [tex]f'(x)=-\dfrac2{x^3}[/tex])
