As you said, let's use the definition: we know that

#f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}#

But #f(x)=\frac{1}{x^2}#, so

#f'(x)= \lim_{h \to 0} \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}#

Let's deal with the numerator: we have

#\frac{1}{(x+h)^2}-\frac{1}{x^2}=\frac{x^2-(x+h)^2}{x^2(x+h)^2}#

Since #(x+h)^2=x^2+2xh+h^2#, this equation becomes

#\frac{1}{(x+h)^2}-\frac{1}{x^2}=\frac{x^2-x^2-2xh-h^2}{x^2(x+h)^2}#

which is of course equal to

#-\frac{2xh+h^2}{x^2(x+h)^2}#

Now, let's return to the limit defining the derivative, and let's plug these results in, we have

#f'(x)= \lim_{h \to 0} -\frac{2xh+h^2}{h*x^2*(x+h)^2}#

First of all, we can simplify #h#:

#f'(x)= \lim_{h \to 0} -\frac{2x+h}{x^2*(x+h)^2}#

Now, since #h# appears only as an additive term, we can simply ignore it, since it tends to 0. So, we have

#f'(x)= -\frac{2x}{x^2*x^2} = -\frac{2}{x^3}#

which is indeed the derivative we are looking for