continuity of function$f(x,y) = xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ in $(0,0)$? I try to find a value for $f(x,y)=xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}}$ in (0,0) which f to be continuous.
First we must show that the $\lim_{(x,y) \to (0,0)} xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ exits and then do the alignment $f(0,0) = \lim_{(x,y) \to (0,0)} xy\dfrac {x^{2}-y^{2}}{x^{2}+y^{2}}$.
I try different ways to calculate the limit but I fail. I try to split the fraction in function to $ \dfrac {xy}{x^{2}+y^{2}}\times\dfrac{1}{x^{2}-y^{2}} $ or $xy \times \dfrac {x^{2}-y^{2}}{x^{2}+y^{2}} $ to simplify  the calculation of limit but I fail.
I even google this and it led to a website which I comment below and it had used the polar coordinates and I don't want to use polar coordinates.
Thanks.
 A: For any given $\varepsilon>0$ we have
$$\begin{align}
\left|\,xy\,\,\frac{x^2-y^2}{x^2+y^2}\right|&\le \frac12|x^2-y^2|\tag1\\\\
&\le \frac12(x^2+y^2)\\\\
&<\varepsilon
\end{align}$$
whenever $\sqrt{x^2+y^2}<\delta=\sqrt{2\varepsilon}$.

Note in arrving at $(1)$ we used the AM-GM inequality $x^2+y^2\ge 2|xy|$.  

Alternatively, note that 
$$\left|\frac{x^2-y^2}{x^2+y^2}\right|\le 1$$
Then, 
$$\begin{align}
\left|\,xy\,\,\frac{x^2-y^2}{x^2+y^2}\right|&\le |xy|\\\\
&\le \frac12(x^2+y^2)\\\\
&<\varepsilon
\end{align}$$

Finally, a transformation to polar coordinates $(x,y)\mapsto (\rho,\phi)$ reveals
$$\begin{align}
\left|\,xy\,\,\frac{x^2-y^2}{x^2+y^2}\right|&=\rho^2|\cos(\phi)\sin(\phi)\,|\cos^2(\phi)-\sin^2(\phi)|\\\\
&\le \frac12 \rho^2\\\\
&<\varepsilon
\end{align}$$
whenever $\rho<\delta=\sqrt{2\varepsilon}$
A: We have $|x|,|y|\leqslant ||(x,y)||$ and by the triangle inequality $|x^2-y^2|\leqslant x^2+y^2=||(x,y)||^2$. Hence
$$|f(x,y)|=\left\vert xy \frac{x^2-y^2}{x^2+y^2} \right\vert\leqslant  ||(x,y)||^2\to 0$$
as $(x,y)\to (0,0).$ The limit exists and is equal to $0$.
