<s>[INST] Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$ Prove $\lim\limits_{(x,y) \to (0,0)} \frac{xy(y^2-x^2)}{x^2+y^2}=0$
My attempt: For all $\epsilon > 0$ there is a $\delta > 0$ such that $\left | \frac{xy(y^2-x^2)}{x^2+y^2} \right| < \epsilon \rightarrow \sqrt{x^2+y^2} < \delta$.
Note important facts: $ (1) \ \sqrt{x^2} < \sqrt{x^2 + y^2}$ and $ (2) \  x^2 < x^2 + y^2$
$$\left | \frac{xy(y^2-x^2)}{x^2+y^2}\right | = \left | \frac{\sqrt{x^2}\sqrt{y^2}(y^2-x^2)}{x^2+y^2} \right| <  \left| \frac{\sqrt{x^2 + y^2}\sqrt{x^2 + y^2}(y^2-x^2)}{x^2+y^2}\right | = |y^2-x^2| $$
From $(2)$ we can show that :
$$|y^2-x^2| < |x^2+y^2-x^2| = |y^2| < |x^2 + y^2| < \delta^2 $$
Hence let $\epsilon = \delta^2$. Does this relationship work?
 [/INST] You should try with polar coordinates, it gets way easier. Impose:
$$\begin{cases} x=\rho\cos\theta\\y=\rho\sin\theta
\end{cases}
$$
and $(x,y)\longrightarrow(0,0)$ will turn into $\rho\longrightarrow0$. As the numerator goes as $\rho^{4}$ and the denominator goes as $\rho^{2}$ your limit is proven. If $\rho$ simplifies from the function then you can't say anything about the limit.
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