Continuity of this function in the origin Does anyone have any idea on how can I prove if this partial derivative is continuous at the origin:
$f(x,y) = \frac{xy(x^2 + 2y^2)}{(x^2+y^2)^{3/2}}$
Of course I know the definition: for every $\epsilon$ there is a $\delta$ if $ ||(x,y)||$ is less that $\delta$ than $|f(x,y)|$ is less than $\epsilon$.77
I'm finding difficult do work with $|f(x,y)|$
I already tried $|f(x,y| = |\frac{xy(x^2 + 2y^2)}{(x^2+y^2)^{3/2}}| \leq |\frac{x(x^2 + 2y^2)}{(x^2+y^2)^{1/2}}|$ because $|\frac{y}{(x^2+y^2)}|\leq1$
But I'm know kind of stucked in there. Can someone help me or give a hint on how to proceed? Thanks! 

HINT:
Transforming to polar coordinates $(\rho,\phi)$, we have
$$\begin{align}
f(x,y)&=\frac{xy(x^2+2y^2)}{(x^2+y^2)^{3/2}}\\\\
&=\frac{\rho^4\sin(\phi)\cos(\phi)(\cos^2(\phi)+2\sin^2(\phi))}{\rho^3}
\end{align}$$
Then, the limit as $\rho\to0$ is trivially $0$.  If $f(0,0)=0$, then $f$ is continuous at the origin.

If one wishes to proceed without a coordinate transformation, simply note that
$$\begin{align}
\left|\frac{xy(x^2+2y^2)}{(x^2+y^2)^{3/2}}\right|&\le \left|\frac{2xy(x^2+y^2)}{(x^2+y^2)^{3/2}}\right|\\\\
&=\left|\frac{2xy}{\sqrt{x^2+y^2}}\right| \\\\
&\le \frac{x^2+y^2}{\sqrt{x^2+y^2}}
\end{align}$$
