Let $U\subseteq\mathbb{A}^{3}$ be $V\left(xy\right)\setminus V\left(x,z\right)$. I'm trying to show that $U$, which is quasi-affine, has a non finitely-generated global section $\mathcal{O}_{U}\left(U\right)$. So far, I looked at the open covering $D_{U}\left(x\right)\cup D_{U}\left(z\right)$, where $D_{U}$ stands for the points in $U$ in which the argument does not vanish, and used the sheaf condition to show that any global function $f$ is locally given by a polynomial function over $D_{U}\left(x\right)$, and same for $D_{U}\left(z\right)$, say $f\vert_{D_{U}\left(x\right)}=g$ and $f\vert_{D_{U}\left(z\right)}=h$, such that $g-h\in\left(y\right)$ and $g,h\in k\left[Y\right]\cong k\left[x,y,z\right]/\left(xy\right)$.
It seems to me that $x$,$z$, and the functions $a_{1}=\left(0,y\right),a_{2}=\left(y,0\right)$, where the first coordinate stands for the restriction of $a$ to $D_{U}\left(x\right)$, and the second to $D_{U}\left(z\right)$, should generate the global section. I tried to extract more information from the inclusion of affine varieties $D_{U}\left(y\right)\subseteq D_{U}\left(z\right)$, but I still don't seem to achieve anything which leads to contradiction. I'd like to know what I'm missing
Also, it should follow that there is no open embedding of $f:U\rightarrow V$ into an affine variety $V$ such that any map from $U$ to an affine variety $W$ factors through $f$. It's obvious it should follow from the fact the global section is not finitely-generated, but I don't see how to prove this formally. Any input would be appreciated.
Thank you!