You have to prove two things. In order to show that $f:\Bbb Z\to\Bbb N$, you must show that $f(n)\in\Bbb N$ for every $n\in\Bbb Z$. In other words, you must show that for each integer $n$, the number $3n^2+2n+5$ is in $\Bbb N$. It’s clear that $f(n)$ is always an integer: it’s a sum of products of integers. What may not be immediately obvious is that it’s always non-negative and therefore in $\Bbb N$ rather than just in $\Bbb Z$. One easy way to see this is to complete the square:
$\begin{align*} f(n)&=3n^2+2n+5\\ &=3\left(n^2+\frac23n+\frac53\right)\\ &=3\left(\left(n+\frac13\right)^2-\frac19+\frac53\right)\\ &=3\left(n+\frac13\right)^2+\frac{14}3\\ &\ge\frac{14}3\;, \end{align*}$
since $\left(n+\frac13\right)^2$ is always at least $0$.
The other thing that you must prove is that $f$ is one-to-one. There’s a standard approach to this that works nicely in most simple problems: assume that $f(n)=f(m)$ for some integers $m$ and $n$, and prove that this forces $m=n$. That shows that if you start with $m\ne n$, you can never get $f(m)=f(n)$, and hence that $f$ is one-to-one.
So begin by supposing that $n,m\in\Bbb Z$ and $f(n)=f(m)$. This simply means that $3n^2+2n+5=3m^2+2m+5\;,\tag{1}$ and all that you have to do is manipulate $(1)$ algebraically until you can infer that $n=m$. Here’s a hint to get you started: $(1)$ is equivalent to $3\left(n^2-m^2\right)+2(n-m)=0\;,$ and if you factor the lefthand side, this becomes $(n-m)\Big(3(n+m)+2\Big)=0\;.$ If a product of two numbers is $0$, what can you say about the numbers? What does that tell you here about $n$ and $m$?