Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!

Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.

Proof:
We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$
Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$
Let us work out the original sequence:
$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$
We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.
This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.
Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$ 

As you were told in the comments, you must take some $\varepsilon>0$ and then prove that $\left|\frac{3n+5}{2m+6}-\frac{3n+5}{2n+6}\right|<\varepsilon$ if $m$ and $n$ are large enough.
Note that$$(\forall n\in\mathbb{N}):\frac{3n+5}{2n+6}=\frac{3n+9}{2n+6}-\frac4{2n+6}=\frac32-\frac2{n+3}.$$So, take $p\in\mathbb N$ such that $\frac2{p+3}<\frac\varepsilon2$. Then$$m,n\geqslant p\implies\left|\frac{3m+5}{2m+6}-\frac{3n+5}{2n+6}\right|=\left|\left(\frac32-\frac2{m+3}\right)-\left(\frac32-\frac2{n+3}\right)\right|<\varepsilon.$$
