Let $I = \langle WX-Y, X^3-Z \rangle \subset \mathbb Q[W,X,Y,Z]$, and $\le$ denote the lexicographic term ordering on $\mathbb N^4$ such that $W > X > Y > Z$.
I've shown:
i) The $S$-polynomial: $S(WX-Y, X^3-Z) = X^2WX-X^2Y-WX^3+ZW = -YX^2+ZW$,
ii) $G=(WX-Y, WZ-YX^2, X^3-Z)$ is the reduced Gröbnerbasis for $I$ with respect to $\le$,
iii) $Q=(W^2X^2+WX^4)^G = Y^2 + ZY \in \mathbb Q[Y,Z]$ is the unique remainder after polynomial division in several variables by $G$,
iv) $Q \in \mathbb Q[Y,Z]$ (trivial),
v) $W^2X^2 + WX^4 = Q(WX,X^3)$
To prove:
However I must show that for $f \in \mathbb Q[W,X]$ and $Q^{'}=f^G$ (the unique remainder), we have $f(W,X)=Q^{'}(WX,X^3)$ if $Q^{'} \in \mathbb Q[Y,Z]$.
We only need to consider the case in which $Q^{'}=f^G \in \mathbb Q[Y,Z]$. Thus $f \in \mathbb Q[W,X]$ must have the form $f=\sum_{i=m+n} a_iW^mX^n$ since each term must be divisible by either $WX$ or $X^3$, otherwise we have a remainder $\in \mathbb Q[W,X]$ and thus $f^G \notin \mathbb Q[Y,Z]$.
Now we we start by dividing by $WX = in_{\le}(WX-Y)$ until no term is divisible by this divisor.
We have the following possibilities for $a_iW^mX^n$:
$m = n$: multiply $WX-Y$ by $a_i(WX)^{m-1}$ to get $a_1(WX)^m -a_i(WX)^{m-1}Y$ and subtract this quantity from $f$. We thus have a new term in $f$, namely $a_i(WX)^{m-1}Y$ which gradually gets reduced to $a_iY^m$ and substituing $WX$ for $Y$ we have $a_1(WX)^m$.
$m > n$: this is not possible since divisions by $WX$ gradually reduce $a_iW^mX^n$ to $a_iW^{m-n}Y^n$ which cannot be reduced further and thus is part of $f^G$, thus $f^G \notin \mathbb Q[Y,Z]$.
$m < n$: we reduce to $a_iX^{n-m}Y^m$ using the argument from the case $m=n$.
Now we only have terms in $f$ of the form $a_iX^n$ or $a_iX^{n-m}Y^{m}$:
Here we reduce $a_iX^n$ to $a_iX^{n-3}Z$ in $f$ by multiplying $X^3 - Z$ by $a_i X^{n-3}$ and subtracting. And gradually we get $a_iX^rZ^k$ where $n = 3k+r$ and $0 \le r < 3$. If $r \neq 0$ then $a_iX^rZ^k \notin \mathbb Q[Y,Z]$ and thus $f^G \notin \mathbb Q[Y,Z]$. Otherwise we have $a_iZ^k$ and substituting $X^3$ we have original term.
This argument also proves the case for $a_iX^{n-m}Y^{m}$ where $n-m = 3k + r$ and $0 \le r < 3$.
Now if $f^G \in \mathbb Q[Y,Z]$ by the previous cases / argument, we have $f^G(WX, X^3) = f \in \mathbb Q[W,X]$.