By definition 

`f'(x)=lim_(Delta x->0)(f(x+Delta x)-f(x))/(Delta x)`

So for `f(x)=1/x` we have

`f'(x)=lim_(Delta x->0)(1/(x+Delta x)-1/x)/(Delta x)=lim_(Delta X->0)((x-(x+Delta x))/((x+Delta x)x))/(Delta x)=`

`lim_(Delta x->0)(x-(x+Delta x))/((x+Delta x)x cdot Delta x)=lim_(Delta x->0)(-Delta x)/((x+Delta x)x cdot Delta x)=`

`=lim_(Delta x->0)-1/((x+Delta x)x)=-1/(x^2)`

So your solution is `f'(x)=-1/x^2`

This is a special case of `(x^n)'=nx^(x-1)` for `n=-1`.