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You can speed up things by factoring the denominator of $f(x,y)$ and see if for each factor you can find its multiple of the form $1+x^l+y^m$ - these are candidate denominators in the decomposition. Assuming that $1+x^l+y^m$ is irreducible whenever $l,m>0$, each factor must be either of this form, or be a polynomial in just one variable and divide some $1+x^l$ or $1+y^m$.

You can speed up things by factoring the denominator of $f(x,y)$ and see if for each factor you can find its multiple of the form $1+x^l+y^m$ - these are candidate denominators in the decomposition. Assuming that $1+x^l+y^m$ is irreducible whenever $l,m>0$, each factor must be either of this form, or be a polynomial in just one variable and divide some $1+x^l$ $2+x^l$ or $1+y^m$.$2+y^m$.

You can speed up things by factoring the denominator of $f(x,y)$ and see if for each factor you can find its multiple of the form $1+x^l+y^m$ - these are candidate denominators in the decomposition. Assuming that $1+x^l+y^m$ is irreducible whenever $l,m>0$, irreducible, each factor must be either of this form, or be form (up to a polynomial in just one variable and divide some $2+x^l$ or $2+y^m$.constant factor).

You can speed up things by factoring the denominator of $f(x,y)$ and see if for each factor you can find its multiple of the form $1+x^l+y^m$ - these are will give you candidate denominators in the decomposition. Assuming Given that $1+x^l+y^m$ is irreducible, irreducible (over $\mathbb{Q}$ at least), each factor must simply be either of this form (up to a constant factor).form.