Suppose $$ f(x)= a_0+a_1x^{-1}+\mathcal O(x^{-2}),\quad x\to\infty $$ and $$ g(y)= b_0+b_1y^{-1}+\mathcal O(y^{-2}),\quad y\to\infty. $$ How would we quote the error term of the product $f(x)g(y)$ for large $x$ and $y$?
Performing term-by-term multiplication one has $$ \begin{align} f(x)g(y) &= (a_0+a_1x^{-1}+\mathcal O(x^{-2}))(b_0+b_1y^{-1}+\mathcal O(y^{-2}))\\ &= a_0b_0+a_0b_1y^{-1} +a_1b_0x^{-1}+a_1b_1x^{-1}y^{-1} +\underbrace{\mathcal O(x^{-2})+\mathcal O(y^{-2})+\mathcal O(x^{-1}y^{-2}) +\mathcal O(x^{-2}y^{-1})+\mathcal O(x^{-2}y^{-2})}_{\mathcal O(???)}.\\ \end{align} $$ If we suppose $x=y$ the answer is obvious: $$ f(x)g(x) = a_0b_0+(a_0b_1+a_1b_0)x^{-1}+\mathcal O(x^{-2}). $$ This shows that the product is $\mathcal O(x^{-2})$.
I did have (the almost certainly incorrect) thought to define polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ so that we consider the expansion $r\to\infty$: $$ \begin{align} f(x)g(y) &= a_0b_0+a_1b_0x^{-1}+a_0b_1y^{-1}+\mathcal O(r^{-2})\\ &=a_0b_0+a_1b_0x^{-1}+a_0b_1y^{-1}+\mathcal O((x^2+y^2)^{-1}), \end{align} $$ since $r^2=x^2+y^2$. But again...this seems almost certainly incorrect.
EDIT:
In this answer I was able to find a definition for multivariable big-$\mathcal O$ that may be helpful here:
We say $u(x,y)=\mathcal O(v(x,y))$ if $\exists c>0$ such that $$ \lim_{\max\{x,y\}\to\infty}(cv(x,y)-u(x,y))\geq 0. $$
Another related question can be found here.
