Take the sequence $x_n = \frac{2}{(4n+1) \pi}$. Then $x_n \to 0$ so if $f$ was continuous, we would have $f(x_n) \to f(0) = 0$. However, $f(x_n) = 1$ for all $n$, so clearly we can't have $f(x_n) \to 0$ and thus $f$ isn't continuous.
Using $\epsilon, \delta$: set $\epsilon = 1/2$. Then for any $\delta >0$, we can find $n$ so that $x_n = \frac{2}{(4n+1) \pi} < \delta$. Then $\lvert f(x_n) - f(0) \rvert = \lvert f(x_n) \rvert = 1 \ge \epsilon.$ That is, for this specific $\epsilon$, there is no $\delta > 0$ which guarantees $\lvert f(x) - f(0) \rvert < \epsilon$ for all $\lvert x - 0 \rvert < \delta$. Hence $f$ is not continuous at $0$.
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.
Best Answer
By your inequality, the absolute value of the difference is $\lt \epsilon$ if $$\frac{a}{e^{\epsilon}}-a \lt x-a\lt ae^\epsilon -a$$ (we subtracted $a$ from each side of each of your two inequalities). Let $\delta=a\min\left(1-\frac{1}{e^{\epsilon}}, e^\epsilon -1\right)$.
Remark: Actually, $1-\frac{1}{e^{\epsilon}}$ is the smaller of the two, so in effect we are letting that be $\delta$. But we really don't need to bother finding that out: all we need to do is to show there is a $\delta$ that works.