I've been staring at these two for a while and I can't come up with anything concrete to start. Hints on how to begin would be greatly appreciated, full solutions are not necessary (and preferably avoided).
Suppose that $F$ is a field and prove the following:
For all $0\ne a \in F, -a\ne 0$ and $(-a)^{-1}=-(a^{-1})$.
For all nonzero $a,b\in F$ and $ab\ne 0$ and $a^{-1}+b^{-1}=(a+b)(ab)^{-1}$
These appeared similar enough to not warrant two posts. Though, I don't know where to begin so I could very well be wrong. I've been staring at the list of axioms for Fields and can't determine how any of them are useful to begin.
The main thing you want to show is that multiplicative inverses are unique: that is, choose $x \neq 0$ in your field. Then the field axioms guarantee an element $y$ such that $xy = 1$, but you can conclude more strongly that there is a unique such $y$.
Given that, it isn't so bad. Remember that the inverse of $x$ is defined to be the (now unique) element such that $xy = 1$. So in the first case, you want to show that $(-a)^{-1} = -(a^{-1})$. Well, all you have to do is show that $$ (-a) \cdot \big(-(a^{-1})\big) = 1 $$ from the above considerations. Now play with associativity and commutativity a little bit, and you should get what you want.
For the second question, it's even easier. Consider multiplying both sides of the (supposed!) equality by $(ab)$.
To be precise, compute first the left-hand side: $$ \begin{align} (a^{-1} + b^{-1})(ab) &= a^{-1}(ab) + b^{-1}(ab) \\ &= (a^{-1}a)b + b^{-1}(ba) \\ &= b + (b^{-1}b)a \\ &= b + a \\ &= a + b \end{align} $$ where on each line we only use one axiom at a time. Now, do the same for the left-hand side, and verify that the two sides are equal.