I think my question has a top-bottom answer, but as of yet I am not familiar enough with divisors and class groups to be sure of what I am claiming. I also include an "elementary answer" to my question below in spoilers, for which I hope I can receive comments and suggestions for simplification.
I am reading Gathmann's lecture notes and upon reading Example 3.3 (and doing Exercise 2.40) I was wondering how one could show the following
Question: Consider the quadric affine hypersurface $X:=V(xw-yz)\subset\mathbb{A}^4$ (any irreducible quadric in $\mathbb{A}^4$ is isomorphic to $X$ via a linear transformation my answer to this question). How can one show that the codimension 1 subvariety $P:=V_X(x,y)\subset X$ (the $z,w$ plane that happens to lie inside $X$) is not a hypersurface in $X$, i.e. there is no $f\in A(X):=\frac{K[x,y,z,w]}{(xw-yz)}$ such that $P=V_X(f)$?
Exercise 2.40 in the aforementioned lecture notes guides us to showing that $I(P)=(x,y)$ can not be generated by a single element, but one can extend the question as above. This extended question is related to Example 3.3: if $\varphi$ were a global quotient of polynomial functions $\frac{g}{f}$ then I can show that $P=V(f)$ which would be a contradiction to the fact stated in my question. Observe that would we have $V_X(f)=P$, this would not imply that $I_X(P)$ is principal.
Top-down exploration: The approach here is via the class group of an affine variety and I am going to use fact from Hartshorne's book (Ch. II Sect. 6).
Unraveling Exercise 6.5 (b) we have that $\mathrm{Cl}(X)=\mathbb{Z}$ where a generator is for instance the image of $P$ in the class group. My question could be reformulated as "Does this imply that $P$ is not a hypersurface?"
Bottom-up approach:
If $V_X(f)=P$, this would imply equations inside $K[x,y,z,w]$ $$x^n=fg+h(xw-yz)\quad\mathrm{and}\quad y^n=fp+q(xw-yz)$$ where $g,h,p,q\in K[x,y,z,w]$ and $n\in\mathbb{N}, n>0$. We can write any polynomial $F\in K[x,y,z,w]$ as $F=\sum_{i,j}F_{i,j}x^iy^j$ with $F_{i,j}\in K[z,w]$. We define $\Sigma(F)$ the set of $(i,j)\in\mathbb{N}^2$ such that $F_{i,j}\in K^\times$. Observe that if $\Sigma(G)=\emptyset$, then $\Sigma(F+G)=\Sigma(F)$.
Step 1: We ca reduce the above equations and assume that $\deg f=\deg_xf=\deg_yf=d$ and $\deg g=\deg_xg=\deg p=\deg_yp=e=n-d$. Indeed first assume that the homogeneous parts of highest degree $f_d,g_e,h_k$ in $f,g,h$ respectively verify $d+e>n$, so that $f_dg_e=-h_k(xw-yz)$, but then one could factor a $xw-yz$ out of either $f$ or $g$, so we can then assume $d+e\leqslant n$ and if $d+e<n$, then $x^n$ would necessarily appear as a monomial in $h(xw-yz)$ which is impossible since any monomial in the latter will necessarily involve at least two variables. For this same reason we equally have that considered as polynomials in $x$, $f$ and $g$ are of degree $d$ and $e$ and with non-zero constant in front of their monomials $x^d$ and $x^e$ respectively: in general $\deg_xf\leq\deg f$ and the same for $g$, but by re-inserting the $x$ expansions of $f,g$ in the first equation we clearly have that $f_dg_e=x^n$. The same argument goes for $f$ and $p$ with respect to $y$.
Step 2 We can now show that $\Sigma(f)=\{(0,d),(d,0)\}$. Indeed observe first by Step 1 that for any $(i,j)\in\Sigma(f),i,j\leqslant d,\ i+j\leqslant d$ and that $(0,d)$ and $(d,0)$ do indeed belong to $\Sigma(f)$. Assume there is one $(i,j) \in \Sigma(f)$ different from either $(0,d)$ or $(d,0)$, so that $i,j<d$. Considering such a pair $(i,j)$ maximal for the lexicographic ordering of $\mathbb{N}^2$, and $(e,0)\in\Sigma(g)$ being maximal as well, we have in $fg$ a monomial of type $x^{i+d}y^j$ which can not be cancelled by $h(xw-yz)$, yet it does not appear in the LHS of the first equation.
Step 3 But wait a minute. Inside Step 2 we already have a way to find a contradiction. Indeed we in fact have proven that there can not be any $(i,j)\neq(d,0)$ in $\Sigma(f)$, since otherwise there would be monomials different from $x^n$ in the first equation that can not be cancelled, yet we have $(0,d)\in\Sigma(f)$.
