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$k$ is a field, then $k[X,Y,W,Z]$ is a domain, so $XW-YZ$ is a non-zero divisor.

$XW-YZ$ is irreducible, then $(XW-YZ)$ is a prime ideal and $k[X,Y,W,Z]/(XW-YZ)$ is a domain, so $Y^2-XZ$ is a non-zero divisor.

That's the easy part. I'm failing to prove that $Z^2-YW$ is a non-zero divisor in $k[X,Y,W,Z]/(XW-YZ,Y^2-XZ)$.

I tried proving this using the fact that, if it is a zero divisor, then we'll have a polynomial $g\in k[X,Y,Z,W]-(XW-YZ,Y^2-XZ)$ where $$g(Z^2-YW)\in(XW-YZ,Y^2-XZ) \Rightarrow$$ $$g(Z^2-YW)=f_1(XW-YZ)+f_2(Y^2-XZ) \hspace{10mm} f_1,f_2\in k[X,Y,Z,W]$$

and tried to expand and find some contradictions in the powers of $Z$, but it led me nowhere.

user8785084
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    $k[x,y,z,w]/I\cong k[s^3,s^2t,st^2,t^3]\subset k[s,t]$ is of dimenison 2, so height of $I\leq2$. If the generators form a regular sequence, then grade of $I$ will be 3, that's not possible. – user782932 Jun 22 '22 at 18:17
  • @user782932 Oh, ok. I was trying to prove this to solve a diferent problem, well i guess this is not the proper way to do it then. Thanks! – user8785084 Jun 23 '22 at 16:50

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