Derivative Solution 
I understand the first step but at the second step how do they come to negative one in the numerator? Also why do they show the definition of derivative again for step 2? How does this produce a negative one?
 A: Let $h$ denote the change in $x$ (delta x).
The definition of a derivative is $$f^{'}(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$$
What the first step did is determine the numerator : $$f(x+h) - f(x) = -\frac{h}{x(x+h)}$$
Dividing the numerator by $h$: $$\frac{f(x+h) - f(x)}{h} = -\frac{1}{x(x-h)}$$
And this is what's happening at step 2. 
Hope this helps. 
A: You say you understand the first step, so you agree that
$$f(x + \Delta x) - f(x) = \frac{-\Delta x }{x(x + \Delta x)}$$
Notice that this is the top portion of the fraction we're looking for in order to find the derivative (be careful, step 2 is still not the definition of the derivative). Now we divide by $\Delta x, $ so
$$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{\frac{-\Delta x }{x(x + \Delta x)}}{\Delta x} = \frac{-\Delta x }{x(x + \Delta x) \Delta x} $$
Now you can see that the term $\Delta x $ cancels out (but notice there is a minus sign) , so
$$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{-1}{x(x+\Delta x)}$$
Now we are ready to take the limit (from the definition of the derivative)
$$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \lim_{\Delta x \to 0} \frac{-1}{x(x+\Delta x)} = - \frac{1}{x^2}$$
