I have a hard time understanding why we pick the $\delta$'s that we do. Let's say we want to check if $\lim_{x \to a} 3x^2 = 3a^2.$
We would proceed as follows. Given that $h = x -a$, and so $x = h+ a$, $|f(x) - 3a^2| = | 3(h+a)^2 - 3a^2| = |6a+3h||h|.$
I then don't understand how we can pick $\delta = \min\left\{ 1, \dfrac{\epsilon}{6|a| + h}\right\}.$
This is a bit of a tricky topic. It turns out that there are infinitely-many different $\delta$ that would work just fine. Picking one that makes a proof easier is largely a matter of experience, and noting what is needed based on your work so far. Sometimes, we'll need a little trial and error to find a good one.
Without seeing the rest of the proof, I can't see off the top of my head why that particular choice of $\delta$ should make our lives any easier. In fact, since we ought to be choosing $h$ based on $\delta,$ I'm fairly certain that it won't!
Let me walk you through the discovery of a $\delta>0$ that will work, so you can get a feel for how such choices may be motivated.
To start with, fix any $\epsilon>0,$ and observe that we're hoping to find some $\delta>0$ such that, for $0<|x-a|<\delta,$ we have $$\left|3x^2-3a^2\right|<\epsilon.$$ Equivalently, making the substitution $h=x-a,$ we want to find $\delta>0$ such that, for $0<|h|<\delta,$ we have $$\left|3(h+a)^2-3a^2\right|<\epsilon.$$
Important Note: If we can find something larger than $\left|3x^2-3a^2\right|$ that is no greater than $\epsilon$ whenever $0<|h|<\delta,$ then we are done. (Can you see why?)
As you mentioned above, we can rewrite $$\left|3(h+a)^2-3a^2\right|=|6a+3h||h|,$$ or we could further rewrite it as $$\left|3(h+a)^2-3a^2\right|=3|h||2a+h|.\tag{1}$$
Now, regardless of our choice of $\delta>0,$ if we have $|h|<\delta,$ then it follows that $$3|h||2a+h|\le 3\delta|2a+h|.\tag{2}$$
Now, by triangle inequality, we have $$|2a+h|\le|2a|+|h|=2|a|+|h|$$ for all $h.$ Thus, for any $\delta>0$ and any $h,$ we have $$3\delta|2a+h|\le 3\delta\bigl(2|a|+|h|\bigr)=6\delta|a|+3\delta|h|.\tag{3}$$ Since we further required that $|h|<\delta,$ then $$6\delta|a|+3\delta|h|<6\delta|a|+3\delta^2.\tag{4}$$
At this point, it looks like we've made our lives more complicated, since now we have $\delta$ and $\delta^2$! However, note that for $0<\delta\le 1,$ we have $\delta^2\le\delta,$ and so $$6\delta|a|+3\delta^2\le6\delta|a|+3\delta=3\delta\bigl(2|a|+1\bigr).\tag{5}$$
By the work above, we have seen that if $0<\delta\le 1,$ then for any $h$ with $0<|h|<\delta,$ we have $$\left|3(h+a)^2-3a^2\right|\overset{(1)}{=}3|h||2a+h|\overset{(2)}{\le}3\delta|2a+h|\overset{(3)}{\le}6\delta|a|+3\delta|h|\overset{(4)}{<}6\delta|a|+3\delta^2\overset{(5)}{\le}3\delta\bigl(2|a|+1\bigr).$$ In summary, if $0<\delta\le 1,$ then for any $h$ with $0<|h|<\delta,$ we have $$\left|3(h+a)^2-3a^2\right|<3\delta\bigl(2|a|+1\bigr).\tag{$\star$}$$ Observe that the right-hand side of $(\star)$ depends only on $\delta$ and $a,$ and in particular, does not depend on $h$. Recall the Important Note. At this point, all we have to do is choose $0<\delta\le1$ such that the right-hand side of $(\star)$ is no greater than $\epsilon,$ and we're done. Since $2|a|+1>0,$ then the right-hand side of $(\star)$ is no greater than (that is, less than or equal to) $\epsilon$ if and only if $$\delta\le\frac{\epsilon}{3\bigl(2|a|+1\bigr)}=\frac{\epsilon}{6|a|+3}.$$ Therefore, putting $$\delta=\min\left\{1,\frac{\epsilon}{6|a|+3}\right\},$$ we have that $\delta>0,$ $\delta\le1,$ and $\delta\le\frac{\epsilon}{6|a|+3}.$ Now that I've done all that work above, I make the following claim:
Take any $\epsilon>0$ and put $$\delta=\min\left\{1,\frac{\epsilon}{6|a|+3}\right\}.$$ Then for any $h$ with $0<|h|<\delta,$ we have $$\left|3(h+a)^2-3a^2\right|<3\delta\bigl(2|a|+1\bigr)\le\epsilon,\tag{$\heartsuit$}$$ and so $$\left|3(h+a)^2-3a^2\right|<\epsilon.$$
Note the similarity between the $\delta$ I chose and the $\delta$ you mentioned? I suspect that the $\delta$ I chose is what was actually intended, and that a typo was made somewhere along the line.
A major problem with a lot of $\epsilon$-$\delta$ proofs is that they don't make clear just how their $\delta$ was determined. There may be a great deal of work that went into it, but by just declaring a $\delta$ and stating that it works without any motivation, the proof's author makes it seem like they are simply privy to some intuition that you, the reader, haven't got. In reality, they probably just don't want to show you all the nitty gritty of the discovery, and don't want to go through all the work of justifying it, either.
Now that I've shown you a $\delta$ that works, see if you can work it backwards and prove $(\heartsuit).$
