Second Derivatives Continuous but Mixed Partials not Equal I'm working on the following problem:

Consider the $f:\mathbb{R}^2 \mapsto \mathbb{R}$ defined by $$f(x,y) = \frac{xy(x^2-y^2)}{x^2+y^2}$$ for everywhere outside the origin and $f(0,0)=0$ at the origin is. Show that the second derivative of $f$ exists but the mixed partials are not equal at the origin.

My work:
I'm getting for $\nabla f = <\frac{yx^4+4x^2y^3-y^5}{(x^2+y^2)^2},\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}>$. My analysis show that the partials are continuous at zero and take the value $0$ at the origin. I'm having trouble analyzing the second derivative:
$$ A = \begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix} =
\begin{bmatrix}\frac{-4xy^3(x^2-3y^2)}{(x^2+y^2)^3}&\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}\\\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}&\frac{4yx^3(y^2-3x^2)}{(x^2+y^2)^3}\end{bmatrix}
$$
$$\lim_{\|h\|\rightarrow 0} \frac{\nabla f((0,0)+h)-\nabla f((0,0))- A \cdot h\|}{\|h\|} \\ = \lim_{\|h\|\rightarrow 0} \frac{\|\nabla f(h)-\langle f_{xx}h_x+f_{xy}h_y,f_{yx}h_x+f_{yy}h_y\rangle\|}{\|h\|} \\= \lim_{\|h\|\rightarrow 0} \frac{\|\nabla f(h)-\langle\frac{-4xy^3(x^2-3y^2)}{(x^2+y^2)^3}h_x+\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}h_y,\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}h_x+\frac{4yx^3(y^2-3x^2)}{(x^2+y^2)^3}h_y\rangle\|}{\|h\|}$$
When I do a polar conversion on this thing and send $r \rightarrow 0$ (uniform for all $\theta$), I get all my $r$'s canceling out -- leaving a limit dependent on theta. But according to the problem, this ratio should go to zero though. Does anyone know what's going on?
I've seen other problems like this that show partials are not continuous at the the origin. That's different though from just establishing existance of the total derivative, right?
 A: To show that the second derivative exists, all you have to show is that the second partials exist at $(0,0)$.  For example $$
f_{xy}(0,0)=\lim_{y\to0}{\frac{f_x(0,y)-f_x(0,0)}{y}}=\lim_{y\to0}{\frac{-y^5}{y^5}}=-1
$$
I also got $$f_{yx}(0,0)=+1,\text{ and }f_{xx}(0,0)=f_{yy}(0,0)=0$$
but I'm getting sleepy, so no guarantees.  
EDIT $$f_{xx}(0,0)=\lim_{x\to 0}\frac{f_x(x,0)-f_x(0,0)}{x-0}=\lim_{x\to0}\frac{0}{x\cdot x^4}=0\\
f_{yy}(0,0)=\lim_{y\to 0}\frac{f_y(0,y)-f_y(0,0)}{y-0}=\lim_{y\to 0}\frac{0}{y\cdot y^4}=0
$$
EDIT
To round this off, I verified that the second derivative doesn't exist at the origin, and I might as well share my calculations.  We know that if the derivative exists it must be $$\pmatrix{0&1\\-1&0}$$ so that the linear approximation to $\nabla f(x,y)$ at $(0,0$) will be $\pmatrix{y\\-x}$.  So we need to show that $$
\lim_{(x,y)\to(0,0)}\frac{\left\lvert\left\lvert\nabla f(x,y)-(y,-x)^{\top}\right\rvert\right\rvert}{||(x,y)||}
$$ is not equal to $0$.  We will show the limit goes to $\infty$ as $(x,y)\to(0,0)$ along the curve $x=y^2$.  It suffices to show the numerator goes to $\infty.$  Also, since $||(a,b)||<|a|,$ it is enough to show this for the first component.  Substituting $x=y^2$ we have$$
\lim_{y\to0}\frac{y^9+4y^7-y^5}{(y^4+y^2)^2}-y=\lim_{y\to0}\frac{2y^7-2y^5}{y^8+2y^6+y^4}=\infty
$$  
