I was playing with $\lim\limits_{(x,y) \to (0,0)} \frac{x^a y^b}{x^c + y^d}$ for $a,b,c,d ,x,y >0 $ and realised that $\frac{x^a y^b}{x^c + y^d}$ is bounded in every neighbourhood of $(0,0)$ if $c=2a , d=2b$ because the $AM-GM$ inequality so is it the only case where $z:=\frac{x^a y^b}{x^c + y^d}$ bounded ? it is easy to see that eventually $x,y <1$ so if $z$ is bounded for some $a,b,c,d$ then any $a_1>a$ and $b_1>b$ and $c_1<c$ and $d_1<d$ would also work
if $c=2a$ and $d>2b$ then the denominator will be $\frac{x^{a}}{y^b} +\frac{y^{b + \epsilon}}{x^a}$ choose some $0<k<\frac{a}{b+\epsilon }$ $y^{b } = x ^{a-k\epsilon}$ it follows that as $x \to 0$ $ \ z \to \infty $ which means if I replaced one power in the demonstrator with higher value than the $AM-GM$ form then $z$ is unbounded
For $(x, y) = (t^d, t^c)$ with $t > 0$ is $$ \frac{x^a y^b}{x^c + y^d} = \frac{t^{ad+bc}}{2t^{cd}} $$ so that a necessary condition for boundedness at the origin is $ad+bc \ge cd$, i.e. $$ \boxed{ \frac ac + \frac bd \ge 1 }\, . $$ Conversely, if $s = \frac ac + \frac bd \ge 1$ then $$ \alpha = \frac{a}{sc} \, , \, \beta = \frac{b}{sd} $$ are positive with $\alpha + \beta = 1$, and $$ x^a y^b = \left( (x^c)^{\alpha} \cdot (y^d)^{\beta}\right)^s \le \left( \alpha x^c + \beta y^d\right)^s \le \left( x^c + y^d\right)^s $$ by the weighted AM-GM inequality, so that $\frac{x^a y^b}{x^c + y^d}$ is bounded near the origin.
Remark: In the same way one can show that $$ \lim_{\substack{(x,y) \to (0,0) \\ x> 0, y > 0}} \frac{x^a y^b}{x^c + y^d} = 0 $$ if and only if $\frac ac + \frac bd > 1$.