Note that $\mathbb{Z_6}$ is not an integral domain, and you also have the requirement that $f(p)$ should be non zero.

Note that $\mathbb{Z}/p \mathbb{Z}$ is a field. Thus, every nonzero element is a unit.

@SeverinSchraven the book's hint states no, and the example is a natural homomorphism from $\mathbb{Z}\to\mathbb{Z}_p$

@Mark so $𝑝$ must be prime, and thte counter example would be $𝑓:\mathbb{Z}\to\mathbb{ℤ}_7,$ defined by $𝑓(𝑎)=[𝑎]_7.$ We take $𝑎=𝑏=2?$

@SeverinSchraven what about I have something like $f:\mathbb{R}\to\mathbb{C}[x],$ defined by $f(x)=2x^2+4?.$ I think the question wants an example in integral domain that is not polynomial related. But I am having a difficult time coming up with an example that is not in the context of polynomials related.

@Seth I am a bit confused by your comment. As I wrote above, every element in $\mathbb{Z}/p \mathbb{Z}$ is not irreducible (it is either zero or a unit). So, yes, the quotient map $\mathbb{Z} \rightarrow \mathbb{Z}/p \mathbb{Z}$ does work as a counterexample if $p$ is prime.

@SeverinSchraven i just have a quick question, in the definition of irreduicble element say $p=ab$ where $p$ is not a unit, and either one of $a, b$ is an associate and the other one is a unit. Is it possible for both $a, b$ to be units. The reason I am asking is that being an associate element does not preclude that associate element being an unit also.

@Seth For me an irreducible element is by definition neither zero nor a unit. However, I do not know what definition you are using.

@SeverinSchraven the definition in my book is: "A nonzero element $p\in R$ is said to be $\textbf{irreducible}$ provided that $p$ is not a unit and the only divisors of $p$ are its associates and the units of $R.$." So to be not irreducible, both divisors of $p$ has not be non units of $R.$

@Seth Yes, that is in line with my previous comment.

@SeverinSchraven don't you meant to say that an irreducible element is by definition either not a zero or is a unit for its divisors? So when we are talking about surjective homomorphism, $f(x)=[x]_p,$ whatever $x$ I plugged into $f$, $f(x)=f(ab)=f(a)f(b)?$ So i am looking at the divisors $a,b?$

Again and for the $n$th time for a very large value of $n$, stop using MathJax for italics and boldface of text that is not part of a formula; that's what mark-up is for. Using MathJax forces font selection that are not device-independent and may yield bad displays. It also slows down processing. Use markup.

@Seth Please reread my previous comments. I cannot be any clearer than I was before. I am saying in a field any element is either zero or a unit and therefore not irreducible. I do not need anything from the homomorphism.

@SeverinSchraven I think there might be something I am not clear about with the map $f(a)=[a]_p,$ where $p$ is prime. let's say $p=7,$ and $x=6$. $6$ is neither irreducible nor a prime, but nevermind the prime part.

@SeverinSchraven When I plug into $f(6)=[6]_7\cdot [1]_7,$ I don't have to deal with factoring $6=2\cdot 3.$ But what if I have $f(6)=f(2\cdot 3)=[2]_7[3]_7?$

$[6]_7$ is a unit. $[2]_7$ is a unit. $[3]_7$ is a unit.

"So to be not irreducible both divisors of $p$ has[sic] to be non units of $R$." That is not the correct negation of "$p\neq 0$, $p$ is not a unit and the only divisors of $p$ are its associates and the units of $R$." The negation is "either $p=0$, or $p$ is a unit, or else $p$ has a divisor that is neither an associate of $p$ nor a unit of $R$." In particular, units are "not irreducibles" automatically, and one does not need to worry about their divisors.

@ArturoMagidin so the only time units comes up in the definition of an irreducible element is in the divsors of the irreducible element. But if the element itself is already an unit, then that element is consider to be not irreducible. Am I correct?

@ArturoMagidin the other possibility for an element $p$ to be not irreducible is that $p$ is not a unit, but none of its factors are units. I think that satisfies the second part of your disjunction in how you corrected my negation? Am I reading it correctly?

Every element of every ring is always divisible by every unit in the ring. Think about it (and I don't need three comments in which you explain it to me).

@ArturoMagidin I want to ask if I am understanding what you are telling me correctly.