If you let $x=r\cos\theta$ and $y=r\sin\theta$ (which is good for checking how the limit looks as you approach the origin from different directions), then the limit becomes $ \lim_{r\to0}\left(r\frac{\cos^4\theta+\sin^4\theta}{\cos^3\theta+\sin^3\theta} \right)=\left\{ \matrix{ \pm\infty&\quad&\theta\to\left(k-\frac14\right)\pi\text{ for }k\in\mathbb{Z}\\\\ \text{undefined}&\quad&\theta=\left(k-\frac14\right)\pi\text{ for }k\in\mathbb{Z}\\\\ 0&\quad&\text{otherwise} } \right. $ and since this depends on $\theta$, the limit does not exist. This is because, on the unit circle, the trigonometric ratio is well-defined except on the two points where $\cos\theta=-\sin\theta$, and at these points, the denominator vanishes while the numerator is positive, producing a ratio that blows up for all $r\ne0$ (and, in our limit, $r$ is never $0$). For other values of $\theta$, the ratio, which can also be represented as $\frac{1+\tan^4\theta}{1+\tan^3\theta}$, is a finite number, so that multiplying it by $r$ scales the result down to $0$ as $r\to0$.