I think the ring $\frac{\Bbb{C}[X,Y,Z]}{(XY-Z)}$ is a UFD.
If we consider the evaluation map $f: \Bbb{C}[X,Y,Z] \rightarrow \Bbb{C}[t^2,t]$ such $f(X)=f(Y)=t$ and $f(Z)=t^2$. Then clearly $(Z-XY)$ is in $\ker f$. Now suppose $p \in \ker f$. Then considering $p$ as a polynomial is $\Bbb{C}[X,Y][Z]$, we have by division algorithm $p=q(Z-XY)+r$, where $q,r \in \Bbb{C}[X,Y][Z] $. If $r \neq 0$, then $\deg r < 1$ in $\Bbb{C}[X,Y][Z]$. So $r \in \Bbb{C}[X,Y]$. Also $f(r)=0$. Now if $r= \displaystyle{\sum_{m,n}c_{(m,n)}X^mY^n}$, then $f(r)=0 \Rightarrow r(t,t)=0 \Rightarrow \displaystyle{\sum_{m,n}c_{(m,n)}t^{m+n}}=0 \Rightarrow c_{(m,n)}=0 \,\forall \, m,n$. So $r=0$. Hence we have $\frac{\Bbb{C}[X,Y,Z]}{(XY-Z)}\approx \Bbb{C}[t^2,t]$.
Now is $\Bbb{C}[t^2,t]$ a UFD?
If so then we are done !
The kernel of your proposed map $\mathbb C[X,Y,Z]\to\mathbb C[t]$ with $X, Y\mapsto t$ and $Z\mapsto t^2$ is not $(Z-XY)$ but $(Z-XY,X-Y)$. However, note that dividing by $(Z-XY)$ just imposes the relation $\overline Z=\overline X\,\overline Y$ in the quotient, which effectively let's us get rid of $Z$:
Consider the ring homomorphism $g\colon \mathbb C[X,Y,Z]\to\mathbb C[X,Y]$ given by $X\mapsto X$, $Y\mapsto Y$ and $Z\mapsto XY$. This is a surjective ring homomorphism with kernel $(Z-XY)$ and hence $$ \mathbb C[X,Y,Z]/(Z-XY) \cong \mathbb C[X,Y]. $$
For any UFD $R$, the polynomial ring $R[X]$ is a UFD as well. Since $\mathbb C$ is a UFD, so is $\mathbb C[X]$ and hence $\mathbb C[X,Y]\cong \mathbb C[X][Y]$ is a UFD as well.