Let's eliminate first the cases $n<0$ and $n\ge2$.
$$f_n(x,y)=\frac{x^2y\ .\ x^{2|n|}}{1+y^2x^{2|n|}}\sim x^{2|n|+2}y\to 0$$
$$f_n(x,x^n)=\frac{x^2x^n}{x^{2n}+x^{2n}}=\frac{1}{2}x^{2+n-2n}=\frac{1}{2}x^{2-n}\to \infty$$
- $n=2$ has multiple limits, so doesn't converges
$$f_2(0,y)=\frac{0}{0+y^2}=0\to 0$$
$$f_2(x,x)=\frac{x^3}{x^4+x^2}\sim\frac{x^3}{x^2}\sim x\to 0$$
$$f_2(x,x^2)=\frac{x^2x^2}{x^4+x^4}\to\frac{1}{2}$$
So let's have now $n\in[0,2[$
We loose no generality in writting $y=x^ng(x,y)$, the function $g$ being arbitrary.
$$f(x,y)=\frac{x^2x^ng(x,y)}{x^{2n}+x^{2n}g(x,y)^2}=x^{2-n}\frac{g(x,y)}{1+g(x,y)^2}$$
According to this study : range of $\frac{x}{x^2+1}$
We can say that $|\frac{g}{1+g^2}|<1$
Thus $|f(x,y)|\le|x^{2-n}||\frac{g}{1+g^2}|<|x^{2-n}|\to 0$
$\color{teal}{Conclusion :}$
$\lim\limits_{x,y\to 0}f(x,y)$ exists for $n\in]-\infty,2[$ and this limit is $0$.
Of course we can also do this directly by setting $u=\frac{y}{x^n}$.
$$f_n(x,y)=\frac{x^2y}{x^{2n}+y^2}=\frac{x^2x^n\frac{y}{x^n}}{x^{2n}(1+\frac{y^2}{x^{2n}})}=x^{2-n}\frac{u}{1+u^2}$$
And since $|\frac{u}{1+u^2}|\le\frac 12$ is bounded, $f_n(x,y)\to 0$ for $n<2$.
Also since we can choose $y$ so that this expression take any desired value in $[-\frac12,\frac12]$ then $f_n(x,y)$ cannot converge for $n\ge 2$.
Note: This is because it has multiple limits. And for $n>2$ the divergence to $\infty$ is not even required, just suffice to say that $-\infty$, $0$, $+\infty$ are all possible limits.