Taking the derivative of $\frac1{x} - \frac1{e^x-1}$ using the definition Given $f$:
$$
f(x) = \begin{cases}
        \frac1{x} - \frac1{e^x-1} & \text{if } x \neq 0 \\
        \frac1{2}                 & \text{if } x = 0
\end{cases}
$$
I have to find $f'(0)$ using the definition of derivative (i.e., limits). I already know how to differentiate and stuff, but I still can't figure out how to solve this. I know that I need to begin like this:
$$
f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{\frac1{h} - \frac1{e^h-1}-\frac1{2}}{h}
$$
But I don't know how to do this. I feel like I should, but I can't figure it out. I tried distributing the denominator, I tried l'Hôpital's but I get $0$ as the answer, while according to what my prof gave me (this is homework) it should be $-\frac1{12}$. I really don't know how to deal with these limits; could someone give me a few tips?

A good strategy in such problems is to massage the problem into recognizable limits. (EDIT: I like this approach mainly because it avoids Taylor expansion and l'Hôpital's rule. This is, however, not the simplest approach.) 
We can "simplify" given function as follows:
$$
\begin{eqnarray*}
\frac{\frac{1}{x} - \frac{1}{e^x - 1} - \frac{1}{2}}{x}  
&=& 
\frac{(2-x)(e^x - 1) - 2x}{2x^2 (e^x - 1)}
\\ &=&
\frac{\color{Blue}{(2-x)}(e^x - 1)- \color{Blue}{(2-x)} \frac{2x}{2-x}}{\color{Red}{2} \ \color{Magenta}{x^2} \color{Green}{(e^x - 1)}}
\\ &=&
\frac{\color{Blue}{2-x}}{\color{Red}{2}} \cdot \frac{\color{Magenta}{x}}{\color{Green}{e^x - 1}} \cdot \frac{e^x - 1 - \frac{2x}{2-x}}{\color{Magenta}{x^3}}. \tag{1}
\end{eqnarray*}
$$
The first two factors both approach $1$ as $x \to 0$. Let us concentrate on the third factor. By Taylor expansion (or the formula for summing a geometric series), we have (for $|x| < 1$),
$$
\begin{eqnarray*}
\frac{2x}{2-x} 
= 
\frac{x}{1 - x/2} 
&=& 
x + x \left(\frac{x}{2} \right)  +x \left(\frac{x}{2} \right)^2 + x \left( \frac x 2 \right)^3 + \cdots
\\ &=& 
\color{Red}{x + \frac{x^2}{2}}  + \color{Blue}{x \left(\frac{x}{2} \right)^2 + x \left( \frac x 2 \right)^3 + \cdots}
\\ &=& 
\color{Red}{x + \frac{x^2}{2}}  + \color{Blue}{x \left(\frac{x}{2} \right)^2 \frac{1}{1 - \frac x 2}} \quad\quad \text{(summing the GP)}
\\ &=&
\color{Red}{x + \frac{x^2}{2}}  \color{Blue}{+\frac{x^3}{2(2-x)}}. \tag{2}
\end{eqnarray*}
$$ 
(Though I used infinite GPs to obtain the final expression, one could verify it directly as well. In particular, the two expressions are equal for all $x \neq 2$, not just $|x| < 1$.) Plugging $(2)$ in $(1)$, we have
$$
\begin{eqnarray*}
\frac{e^x - 1 - \frac{2x}{2-x}}{x^3}
&=& 
\frac{e^x - 1  \color{Red}{- x - \frac{x^2}{2}} \color{Blue}{-\frac{x^3}{2(2-x)}}}{x^3}
\\ &=&
\frac{e^x - 1 \color{Red}{- x - \frac{x^2}{2}}}{x^3} \color{Blue}{-\frac{1}{2(2-x)}}
\end{eqnarray*}
$$
Once again, the second term has an easy limit of $\frac14$. The first term
$$
\frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}
$$
is also a standard limit. This limit can be evaluated using, say, the l'Hôpital's rule or using the Taylor expansion of $e^x$; it's value is $\frac{1}{3!}$. Plugging in both these limits, we can get the final answer to be
$$
\frac{1}{3!} - \frac{1}{4} = -\frac{1}{12}.
$$

Bonus! If you wish to avoid Taylor expansion and l'Hôpital's rule even further, I will mention an "elementary" ways to evaluate limits such as 
$$
\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} \quad\text{and}\quad \lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3} 
$$
assuming that these limits exists! (I stress that this is not a complete proof; yet I present it because I find the technique interesting.)
I will show the idea for the first limit, and leave the second one as an exercise. Suppose $A \stackrel{\text(def)}{=} \lim \limits_{x \to 0} \frac{e^x - 1 - x}{x^2}$ exists. Then $e^x = 1 + x + A x^2 + o(x^2)$. Therefore, by squaring: $$e^{2x} = (1 + x + A x^2 + o(x^2))^2 = \color{Red}{1} + \color{Blue}{2x} + \color{DarkGreen}{x^2 (1+2A)} + o(x^2) .$$ On the other hand, making the substitution $x \to 2x$ in the definition of the limit, we have $e^{2x} = \color{Red}{1} + \color{Blue}{2x} + \color{DarkGreen}{4A x^2} + o(x^2)$. 
Equating the dominant terms in these two expressions, we must have $\color{DarkGreen}{4A} = \color{DarkGreen}{1 + 2A}$, which gives $A = \frac{1}{2}$. 
