Continuity of piecewise $f$ and its partials, $f(x,y)= \frac{xy(x^2-y^2)}{x^2+y^2}$ for $x,y\neq 0$ 
Let $f: \mathbb{R}^2 \to \mathbb{R}$ and $$f(x,y) = \left\{
\begin{array}{ll}
      0, & (x,y)=0 \\
      \frac{xy(x^2-y^2)}{x^2+y^2}, & (x,y) \neq0 \\
\end{array} 
\right.$$ Determine if $f, \partial_xf$ and $\partial_yf$ are continuous.

So in $\mathbb{R^2}_{\ne 0}$ we have that $f$ is a rational function hence continuous. For the case $(x,y) = 0$ I managed to get the following:
Let $\varepsilon>0$ $$|f(x,y)-f(0)|=\left|\frac{xy(x^2-y^2)}{x^2+y^2}\right| \leqslant \frac{|xy||(x^2+y^2)|}{x^2+y^2} = |xy| = (xy)^2 < \varepsilon$$ when $xy < \sqrt{\varepsilon.}$
However for $ \partial_xf = \frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}$ I cannot seem to be able to get anything. Also I'm not sure does the fact this is also a rational function imply continuity at $\mathbb{R^2}_{\ne 0}$?

Hint:
Transform the partial derivatives via polar coordinates. Also there's an error in your  partial derivative w.r.t. $x$:
$$\frac{\partial f}{\partial x}=\frac{y(x^4+4x^2y^2-y^{\color{red}4})}{(x^2+y^2)^2}=\frac{r^5(\cos^4\theta+4\cos^2\theta\sin^2\theta-\sin^4\theta)}{r^4}=r(\cos2\theta+\sin^22\theta),$$
therefor it clearly tends to $0$, and similarly for the other partial derivative.
