Disclaimer: This is a question from a book on functions of several variables. The answer is in the back of the book.
I have a function
$f(x, y) = \begin{cases} \frac{xy(x^2 - y^2)}{x^2 + y^2} &\text{for}\ (x,y) \not= (0,0) \\ 0 &\text{for}\ (x, y) = (0, 0) \end{cases}$
I have shown that it is continuous, by converting to polar coordinates and shown that
$\lim_{(r, \theta) \rightarrow (0, 0)} = 0$
I have also done the partial derivative and they too are continuous, so the function is $C^1$.
Using
\begin{align} \lim_{h \rightarrow 0} \frac{\frac{\partial f}{ \partial y}(h, 0) - \frac{\partial f}{\partial y}(0, 0)}{h} = 1 \end{align}
So we can see that $\frac{\partial^2 f}{\partial y \partial x}(0, 0) = 1$ and likewise we see that $\frac{\partial^2 f}{\partial y \partial x}(0, 0) = -1$.
My question is now: How do we prove that the function is not $C^2$? The answer at the back of the book just states that it is not, however as far as the definition of $C^2$ goes we just have to prove that the mixed derivatives exists and are continuous, which I believe that they are? What am I missing? I get that they are not the same, and that a $C^2$-function have equal mixed derivatives, however the converse isn't true.