So the full problem was:
Consider $R=\mathbb Q[x,y,z]/(x-xyz)$. Prove that $x$ and $xy$ divide each other in $R$ but that they are not associates. In other words, there is no unit $u\in R$ so that $x=xyu$.
I already proved they divide each other but I need to prove no unit exists.
EDIT: After a helpful observation and correction from user26857, I have made significant changes in the fourth paragraph below (now the fifth!), related to the units of $A$ and those of $B$. I’m afraid that my proof is now a bit messier.
As I suggested in my comments, the key to this question is to identify precisely what the units of $R$ are. And below, I mean to persuade you (and myself) that the only units of $R$ are the constants.
Consider the rings $A=\Bbb Q[x,y,z]/(x)$ and $B=\Bbb Q[x,y,z]/(1-yz)$. Of course you see immediately that $A\cong\Bbb Q[y,z]$ and $B\cong\Bbb Q[x,y,1/y]$. Furthermore, we have a map into $A\oplus B$: the underlying set is the product set $A\times B$, with both addition and multiplication proceeding coordinatewise. Its units are quantities of the form $(\alpha,\beta)$ with $\alpha\in A^*$ and $\beta\in B^*$: both coordinates have to be units.
What’s the kernel of the map $\pi:\Bbb Q[x,y,z]\to A\oplus B$? It’s the intersection of the two separate kernels, thus $(x)\cap(1-yz)$. But the polynomials $x$ and $1-yz$ are nonassociate irreducibles in the UFD $\Bbb Q[x,y,z]$, and anything divisible by both is divisible by their product. Thus the kernel of $\pi$ is our ideal $(x-xyz)$.
The map $\pi$ is not onto, but this does not deter us: we now have $R\subset A\oplus B$, and any unit of $R$ must be mapped via $\pi$ to a unit of $A\oplus B$. You see that a $\Bbb Q$-polynomial $g(x,y,z)$ goes to a unit of $A$ if and only if it’s of form $\mu+xh(x,y,z)$ with $\mu\ne0$. But the units of $B\cong\Bbb Q[x,y,1/y]$ are things of the form $\lambda y^m$ with $\lambda$ nonzero in $\Bbb Q$. Now our $\mu+xh(x,y,z)$ goes to $\mu+xh(x,y,1/y)$ in $B$. But this is not a unit there unless $h=0$. With this determination, you see immediately that $x$ and $xy$ are not associate, because they’re not related by a constant multiplier.
As the answer in the linked thread shows us, there is no need to find the units of $R$ in order to solve the problem. But if we want to do this, then the nicest way I know is the following (instead of $\mathbb Q$ let's consider an arbitrary field $k$):
There is an injection $$R \hookrightarrow k[X,Y,Z]/(X)\times k[X,Y,Z]/(1-YZ)\simeq k[Y,Z]\times k[X,Y,Y^{-1}].$$ This induces an injection $$R^\times \hookrightarrow (k[Y,Z]\times k[X,Y,Y^{-1}])^\times\simeq k^\times \times k^\times Y^{\mathbb Z}.$$ Therefore a unit in $R$ is the image of a polynomial $f\in k[X,Y,Z]$ that is simultaneously a nonzero constant modulo $X$, and a nonzero constant times $Y^n$ modulo $1-YZ$. Being a unit modulo $X$ means $f=u+Xg$ for some $g\in k[X,Y,Z]$, $u\in k^\times$ while the second condition implies $u+Xg\equiv vY^n\bmod(1-YZ)$ for some $v\in k^{\times},n\in\mathbb Z$. By sending $X$ to $0$ we deduce that $n=0$ and $u=v$, so $Xg\equiv 0\bmod(1-YZ)$. It follows that $1-YZ\mid g$, and thus $f=u+X(1-YZ)g$ which gives $f=u$ in $R$, and we are done.
