I calculated it as follows, if we use the polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ as $r$ goes to zero, we get $ \frac{((r\cos\theta)^2+(r\sin\theta)^2)^2}{(r\cos\theta)^2 - (r\sin\theta)^2} $ , which would simplify to:
$$ \frac{((r\cos\theta)^2+(r\sin\theta)^2)^2}{(r\cos\theta)^2 - (r\sin\theta)^2} = \frac{r^4}{r^2(\cos\theta)^2 - r^2(sin\theta)^2} = \frac{r^2}{(\cos\theta)^2 - (\sin\theta)^2}$$, as $r$ goes to $0$. The limit would be zero when $(\cos\theta)^2 - (\sin\theta)^2 \neq 0$ and the limit is undefinied when $(\cos\theta)^2 - (\sin\theta)^2 = 0$, so the limit is undefined when $(\cos\theta)^2 = (\sin\theta)^2$, solving the equation we find $\theta \in \{ \pi/4 , - \pi/4, 3 \pi / 4, 5 \pi / 4 \}$.
So the limit is $0$ except when $\theta \in \{ \pi/4 , - \pi/4, 3 \pi / 4, 5 \pi / 4 \}$ the limit is undefined.
Is this solution correct?
$$\lim_{(x,y)\to(0,0)} \frac{(x^2+y^2)^2}{x^2 - y^2}=\lim_{(x,y)\to(0,0)} \frac{(x^2+y^2)^2}{(x-y)(x+y)} $$ Consider the path $y=x-x^3$ and note that along this path $y\sim x$ as $x\sim 0$: $$\frac{(x^2+y^2)}{(x-y)(x+y)}=\frac{(x^2+(x-x^3)^2)^2}{(x-x+x^3)(x+x-x^3)}\sim \frac{4x^4}{2x^4}\to 2$$ Thus the limit does not exist.