$(\varepsilon, \delta)$ for continuity of a multivariable function 
Define $f : \Bbb R^2 \to \Bbb R^2$, $$f(x,y) = \begin{cases} 
      0 & (x,y) =  0 \\
      \frac{xy(x^2-y^2)}{x^2+y^2} & (x,y) \ne 0
   \end{cases} $$ Determine if $f$ is continuous.

At $\Bbb R^2_{\ne 0}$ we have that $f$ is continuous. In order to see if it's continuous at the origin I was approaching it using epsilon-delta. We have that $|f(x,y) - f(0,0)| < \varepsilon$ whenever $\sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2} < \delta$.
Now $$|f(x,y)-f(0,0)| = \frac{|xy||x^2-y^2|}{|x^2+y^2|} \leqslant \frac{|xy||x^2+y^2|}{|x^2+y^2|} =|xy|.$$
However I'm losing $x^2+y^2$ here since that's what I would need for $\delta$. What am I missing in my approach here? Is there another way I could bound this?

For all $(x,y)\in\mathbb{R}^2$ we have
$$
\left\lbrace
\begin{align*}
&2|xy|\le x^2+y^2\\[4pt]
&|x^2-y^2|\le x^2+y^2\\[4pt]
\end{align*}
\right.
$$
hence for all $(x,y)\in\mathbb{R}^2{\setminus}\{(0,0)\}$ we have
$$
2|f(x,y)|
=
\frac
{(2|xy|)(|x^2-y^2|)}
{x^2+y^2}
\le
\frac
{(x^2+y^2)(x^2+y^2)}
{x^2+y^2}
=
x^2+y^2
$$
which approaches $0$ as $(x,y)$ approaches $(0,0)$.
