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?
 A: 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.
A: Your last line of inequalities should be changed to $|y^{2}-x^{2} |= |y^{2}|+|x^{2}|\leq y^{2}+x^{2}$ without the intermediate steps (which are wrong).
If $x=\sqrt 3$ and $y=1$ the inequality $|y^{2}-x^{2} |\leq y^{2}$ is false.
A: Your last step is not correct as for example take $|2^2-5^2|\lt |5^2+2^2-5^2|$, which is clearly not true.
You may proceed like this also: 
For any $\epsilon\gt 0$,
$\begin{align}|\frac{xy(y^2-x^2)}{x^2+y^2} |& = |\frac{\sqrt{x^2}\sqrt{y^2}(y^2-x^2)}{x^2+y^2} | \\& \lt  | \frac{\sqrt{x^2 + y^2}\sqrt{x^2 + y^2}(y^2+x^2)}{x^2+y^2} | = y^2+x^2\lt \epsilon \end{align}$
Now choose $\delta =\sqrt \epsilon$ to prove your claim.
A: Otherwise it is ok. I would write it however a bit different:
\begin{align}
&(x-y)^2\ge 0~\implies~|xy|\le\frac{x^2+y^2}{2}\le x^2+y^2\\
&\left|\frac{xy(y^2-x^2)}{x^2+y^2}\right|\le|y^2-x^2|\le x^2+y^2\le\delta^2=\varepsilon
\end{align}
when choosing $\delta=\sqrt{\varepsilon}$.
