The answer is yes to all $(x,y)$. Claim: if $t>10^6$ then $f(g(t^3))-g(f(t^3))\in [1,1.001]$ Proof: \begin{align*} f(g(t^3))&=f(t^3+t)\\ &=(t^3+t)+3(t^3+t)^{\frac 23} \\ &= t^3+t+3(t^6+2t^4+t^2)^{\frac 13}\\ &= t^3+t+3(t^2+\frac 23-\frac 19 t^{-2})\\ &= t^3+3t^2+t+2-O(t^{-2}) \end{align*} Meanwhile, \begin{align*} g(f(t^3))&= g(t^3+3t^2)\\ &= t^3+3t^2+\left(t^3+3t^2\right)^{\frac 13}\\ &= t^3+3t^2+(t+1-\frac 1t)+O(t^{-2})\\ &= t^3+3t^2+t+1-\frac 1t + O(t^{-2}) \end{align*} Now, the key idea is to stuff $n,t,k$ such that if I construct $$f^n(g^n(t^3)) > b_k > f^n(g^n(t^3)) - n^2 > g^n(f^n(t^3))$$ Where $t$ is a fixed term in the sequence $a_i$ such that $|t-n^{\frac 32}| < 1$, and $t,n>10^{18}$. We can do this by making sure that $b_k$ increments by less than $n^2$ when the inequalities hold (i.e. $b_k<n^6$ because we can just apply $g$.) Note $t^3<f^n(g^n(t^3)) < f^{2n}(t^3)<(t+2n)^3$, since $f(t^3)<(t+1)^3$. Therefore, I have $b_k=g(b_{k-1})$ for every $k$ and since $b_k^{\frac 13}$ increases by at most $\frac{1}{2b_k^{\frac 13}}$ every time, I can guarantee that there exists $m$ such that $b_{m-1} < t^3 < b_m$ and $$b_m^{\frac 13} < b_{m-1}^{\frac 13} + \frac{b_{m-1}^{\frac{-1}{3}}}{2} < t+\frac{1}{2t}$$ The increase in $b_m$ is therefore $b_{m-1}^{\frac 13} < t+2n <n^2$ since $n>t^{\frac 23}-1$ Now, having constructed $k,n,t$ such that $$f^n(g^n(t^3)) > b_k > f^n(g^n(t^3)) - n^2 > g^n(f^n(t^3))$$we apply discrete continuity. Set $p$ such that $a_p=t^3$. Consider a string $S=fffff\cdots fgggg\cdots g$, which has $n$ f's followed by $n$ g's. Each time I swap an adjacent $f,g$ in $S$. Let $s_i$ be the $i$th character of $S$, then we set $a_{p+j}=f(a_{p+j-1})$ if $s_j=f$ and $g(a_{p+j-1})$ otherwise. We can see $a_{p+2n}$ decreases by a value between $[1,1.01]$ with each transposition, so at one point $$|a_{p+2n}-b_k|<1$$ Must hold.