When is this rational fraction bounded? Let $\Omega=(0,\infty)^2$. For $\alpha \gt 0, \beta \gt 0$, define a function
$f_{\alpha,\beta}$ on $\Omega$ by putting
$$
f_{\alpha,\beta}(x,y)=\frac{(xy)^{\alpha}(1+x^2)(1+y^2)}{\big(xy(x+y)+1\big)^{\beta}}
$$
For which pairs  $(\alpha,\beta)$ is $f_{\alpha,\beta}$ bounded ?
Putting $x=y=t$ and letting $t\to \infty$, we see that $3\beta \geq 2(\alpha+2)$ is a necessary condition. I don’t know if it is sufficient.
 A: The necessary and sufficient condition is $3\beta\ge 2\alpha+4$ and $\alpha\ge 2$.
When $x,y\ge 1$, you have seen $3\beta\ge 2\alpha+4$ is necessary. In this case, it is also sufficient. Note that $xy(x+y)+1\ge 2(xy)^{\frac{3}{2}}$, and  when $x,y\ge 1$,  $1+x^2\le 2x^2$, $1+y^2\le 2y^2$. It follows that
$$f_{\alpha,\beta}(x,y)\le 2 (xy)^{\alpha+2-\frac{3}{2}\beta}\le 2.$$
When $x\le 1$ or $y\le 1$, an additional condition is $\alpha\ge 2$. By symmetry, we may assume $x\le 1$. When $y\le 1$, it it easy to see $f_{\alpha,\beta}(x,y)\le 4$, so let us also assume $y\ge 1$. Then $1\le 1+x^2\le 2$, $y^2\le 1+y^2\le 2y^2$ and $xy^2\le xy(x+y)\le 2xy^2$, and hence
$$\frac{x^\alpha y^{\alpha+2}}{(2xy^2+1)^\beta}\le f_{\alpha,\beta}(x,y)\le \frac{4x^\alpha y^{\alpha+2}}{(xy^2+1)^\beta}.$$
Let $z=xy^2$. If $z=1$, 
$$\frac{x^\alpha y^{\alpha+2}}{(2xy^2+1)^\beta}=3^{-\beta}y^{2-\alpha}, $$
so $\alpha\ge 2$ is necessary. 
Now let us show the sufficiency. If $z\le 1$, then 
$$\frac{x^\alpha y^{\alpha+2}}{(xy^2+1)^\beta} \le x^\alpha y^{\alpha+2}=z^\alpha y^{\alpha-2}\le 1.$$
If $z\ge 1$,
$$\frac{x^\alpha y^{\alpha+2}}{(xy^2+1)^\beta}\le x^{\alpha-\beta}y^{\alpha+2-2\beta}=z^{\alpha-\beta}y^{2-\alpha}.$$
When $\alpha\ge \beta$, since $x\le 1$, $y\ge 1$ and $3\beta\ge 2\alpha+4$,
$$x^{\alpha-\beta}y^{\alpha+2-2\beta}\le y^{\frac{1}{2}(2\alpha+4-3\beta)}\le 1;$$
when $\alpha\le \beta$, since $z\ge 1$, $y\ge 1$ and and $\alpha\ge 2$,
$$z^{\alpha-\beta}y^{2-\alpha}\le 1.$$
A: The denominator of your function cannot become zero for any values of $x$ and $y$, so that your function cannot have a singularity. It remains to analyse the behaviour of your function in the far-field. We consider all sequences $(x_n,y_n)$ where $x_n^2+y_n^2\to\infty$. In this set of sequences, we distinguish between the sequences where either $x_n\to0$ or $y_n\to0$ (Type I), and all other sequences (Type II). For sequences of Type I without loss generality $x_n\to0$, then we can write
$$ f_{\alpha,\beta}(x_n,y_n)\sim\frac{x_n^\alpha y_n^{2+\alpha}}{(x_ny_n^2+1)^\beta}\,.$$
In the following we show that the right hand side is bounded if and only iff $\alpha\ge2$ and $\beta\ge\alpha$:
(...)
For all other sequences, we introduce polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ where $0<\theta<\pi/2$. Our function in polar coordinates is
$$f_{\alpha,\beta}(r,\theta)=r^{2\alpha+4-3\beta}\frac{(\cos\theta\sin\theta)^\alpha(\frac1{r^2}+\cos^2\theta)(\frac1{r^2}+\sin^2\theta)}{(\cos\theta\sin\theta(\cos\theta+\sin\theta)+\frac1{r^3})^\beta} $$
Our function is bounded for all sequences of Type II if and only if
$$ 2\alpha+4-3\beta\le0$$
