The question is
For any quasi-affine variety $X$, there exists $n, m$ such that $X$ is isomorphic to a closed subvariety of the quasi-affine variety $\mathbb{A}^n\setminus \mathbb{A}^m$.
Intuitively I know that given an quasi-affine variety $X$, I need to take $n,m$ "big enough" so that I can embed $X$ linearly (maybe not?!) into $\mathbb{A}^n\setminus \mathbb{A}^m$. But how do you do that? I tried some examples and came up empty.
I think if somebody can give me an example of how to show this for a specific quasi-affine variety (which is not completely trivial). I think a good example which is general enough can be:
$X=Z\cap U$ where $Z=\{(x,y)\in \mathbb{A}^2|y^3-x^2=0\}$ and $U$ is complement of twisted cubic curve $U=\{(x,y)\in \mathbb{A}^2|y^2-x^2(x+1)\neq 0\}$.
I specifically tried working this out but after spending hours playing around, drawing stuff and staring at it, I realized I don't really know what I'm doing and don't know really where to begin.