There is no "sure fire" way of proving continuity of a function. However, the steps are usually a bit backward to what the actual definition is. That is, the definition says that $f$ is continuous at $a$ if for each $\epsilon>0$, there exists $\delta >0$ such that if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.
We start the proof by taking an arbitrary $\epsilon > 0$. However, we then usually do not magically think of a $\delta$ that would fit. What we typically try to do is simplify the expression $|f(x)-f(a)|$ and prove that it is "small", keeping in the back of our mind that we can always make $|x-a|$ "small".
In your case, for example, we can prove first that $|f(x)-f(a)|=|\frac{x-a}{\sqrt x + \sqrt a}|$. Now, we want to ask ourselves: If $|x-a|$ is small, is this expression also small?
This is done by trying to find a small upper bound for the expression that will hold whenever $|x-a|<\delta$. In your particular case, this is fairly simple, since we know that the expression is smaller than $\frac{|x-a|}{\sqrt a}$, and this is smaller than $\epsilon$ if $\delta$ is set small enough.
Now, once we did these steps, we take a step back, and think about what we just did.
Looks like we found our $\delta$, and we have to set our $\delta=\epsilon\sqrt a$.
Hmm, is this OK? Well, as long as $a>0$, we know that $\delta>0$, and we have proven that if $|x-a|<\delta$, then $|f(x)-f(a)|<\frac{\delta}{\sqrt a}$, ,which is equal to $\epsilon$, so it looks like we are almost done.
Almost, because we can see that if $\sqrt a=0$, then our proof does not work! We need to do that part separately. If $a=0$, then $|f(x)-f(a)|=|\sqrt x|$. We still need to prove that if $|x-0|$ is small, then $|\sqrt x|$ is also small.
Best Answer
Let $\epsilon>0$. We want to pick a $\delta$ such that $|x-(-1)|=|x+1|<\delta\implies|f(x)-f(-1)|=|f(x)|<\epsilon$.
Note that
$$ |f(x)|=\bigg|\frac{3x^{2}-2x-5}{2x+3}\bigg|=\bigg|\frac{(3x-5)(x+1)}{2x+3}\bigg|=\bigg|\frac{3x-5}{2x+3}\bigg||x+1|. $$
We are free top choose $\delta$ as we wish, and, in doing so, we will have a handle on making the $|x+1|$ term above as small as we want. To help us minimize $\big|\frac{3x-5}{2x+3}\big|$, let's see what happens if we would ensure $\delta <\frac{1}{3}$.
If $|x+1|<\frac{1}{3}$, then $-\frac{1}{3}<x+1<\frac{1}{3}$. This implies that
$$ -1<3x+3<1\implies-9<3x-5<-7<9\implies|3x-5|<9 $$
and
$$ -\frac{2}{3}<2x+2<\frac{2}{3}\implies\frac{1}{3}<2x+3<\frac{5}{3}\implies\frac{1}{3}<|2x+3|. $$
Thus,
$$ \bigg|\frac{3x-5}{2x+3}\bigg|<\frac{9}{1/3}=27. $$
So, if $\delta<\frac{1}{3}$, it follows that $|f(x)|<27|x+1|$. So, if in addition to this, $\delta<\frac{\epsilon}{27}$, we get that $|f(x)|<27\cdot\frac{\epsilon}{27}=\epsilon$.
Hence, we could choose $\delta=\min\{\frac{1}{3},\frac{\epsilon}{27}\}$. Then, if $|x+1|<\delta$, it follows that $|f(x)|<\epsilon$, as desired.