How to calculate second derivative using definition?

Question

I am currently reading Analysis on Manifolds by Munkres and this problem has me flustered.

$f(x, y) = \frac{xy(x^2 - y^2)}{x^2 + y^2}$ and $f(0) = 0 $. Show that $D_1D_2 f$ and $D_2 D_1f$ exist at $0$ but are not equal there?

I can calculate $D_1D_2f$ and $D_2D_1f$ using partial differentiation. Thus \begin{align} D_1D_2 f &= \frac{\partial }{\partial x}\left( \frac{\partial f}{\partial y}\right) = \frac{(x^2 - y^2)(x^4 + 10x^2y^2 + y^4)}{(x^2 + y^2)^3}\\ D_2D_1f &= \frac{\partial }{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{(x^2 - y^2)(x^4 + 10x^2y^2 + y^4)}{(x^2 + y^2)^3} \end{align}

So I know for a fact that this is not the way to go. However, how do I actually calculate $D_1 D_2 f(\bf 0)$ using definition of derivative? I can calculate $D_2 f(\bf 0)$ as follows: \begin{align} D_2 f({\bf 0}) &= \lim\limits_{\lambda \to 0}\frac{f({\bf 0} + \lambda(0 , 1)) - f({\bf 0})}{\lambda} = \lim\limits_{\lambda \to 0} \frac{f(0, \lambda) - 0}{\lambda} \\ &= 0 \end{align}

How do I get a closed form function for $D_1 f$ 0r $D_2 f$ so I can take a second derivative?

Answer 1

You have

$$\begin{cases} \frac{\partial f}{\partial x}(0,y)=-y\\ \frac{\partial f}{\partial y}(x,0)=x \end{cases}$$

From which it follows that:

$$\frac{\partial^2 f}{\partial x \partial y}(0,0)=1\neq −1= \frac{\partial^2 f}{\partial y \partial x} (0,0)$$