Prove that $-(ab) = a(-b)$ 
Prove that $$-(ab) = a(-b)$$

Proof:
We need to show that $$a(-b) + ab = 0$$
which is equivalent to
$$\begin{align}
a(-b)+ ab &= 0 \\
a(-b+b) &= 0 \\
a(0) &= 0 
\end{align}$$
Q.E.D
I would like to have some feedback on the proof. Thanks.
 A: Moving my comment to an answer and expanding a bit.
You're doing the necessary steps to get the proof. Now you just need to write it in reverse order, starting with $a(0)=0$. But is this true? Why?
Really, you're trying to show "For any $a$ and for any $b$, we have ..." that equality.
So the proof could resemble:

Let $a$ be arbitrary. Then by _____, we have $$a(0)=0.$$ Now let $b$ be arbitrary. By _____, we have $-b+b=0,$ and substituting into the above equation, we obtain $$a(-b+b)=0.$$
...

A: Your idea is correct. You just have to understand what you are doing. You know that for $a \in \mathbb{R}$, we have $a.0 = 0$ because $a.0 = a (0 + 0) = a.0 + a.0$. By the uniqueness of the neuter element of the sum we obtain $a.0 = 0$ . Now, for any real number $b$ we have $b-b = 0$. Therefore $0 = a.0 = a (b-b) = ab + a (-b)$. why?
A: I would do the proof as follows:
Let $a, b \in R$, your ring.
From an earlier result we know that:
$a \cdot 0=0$  (so multiplication by the additive identity element results in the additive identity)
If you do not know this, prove it first (hint $0=0+0$).
Next we rewrite this using $b+-b=0$. This gives:
$$a(b+-b)=0.$$
We apply the distributive law:
$$ab + a(-b) =0$$
Now observe that this condition means that $a(-b)$ is the inverse of $ab$. Normally we would write this as $-ab$. By uniqueness of inverses we know that $a(-b)= -ab$ $\square$.
It is best practice to explain your reasoning and which laws/axioms you apply in between steps. Your original proof consisted of symbolic manipulation. You have to think and argue "why" your steps are true.
