<s>[INST] Limit of $f(x, y)$ at $(0, 0)$ I need to show 
$$\lim_{(x, y) \rightarrow (0,0)} \frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} = 0
$$
Not really sure how to go about this without using the epsilon delta definition, which I would prefer not to. Any sort of help is appreciated.
Edit: I do have the inequality:
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} \le \frac{xy(x^2+y^2)}{(x^2+y^2)^{3/2}} = \frac{xy}{(x^2+y^2)} $$
 [/INST] Use polar coordinates: by letting $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have that
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}= \frac{r^4\sin(2\theta)\cos(2\theta)}{2r^3}=\frac{r\sin(4\theta)}{4}.$$
Therefore
$$\left|\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}\right|\leq \frac{r}{4}$$
and the right-hand side goes to $0$ as $r\to 0$.
P.S. Your equality is too "weak" for the job: by taking $y=x$ the right-hand side is $\frac{1}{2}$.
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