Prove at $(0,0)$, $f_{xy}$ and $f_{yx}$ aren't equal

Question

If $f(x, y) =xy(\frac{x^2-y^2}{x^2+y^2})$ when $x$ and $y$ are not simultaneously zero when $x=0,y=0$. Show that at $(0,0),\ f_{xy} \neq f_{yx} $

Answer 1

First find $f_x$ and $f_y$.

For $(x,y) \neq (0,0)$ you have

$$f_x(x,y) = y \frac{x^2-y^2}{x^2+y^2} + xy \frac{2x(x^2+y^2) - 2x (x^2 - y^2)}{(x^2+y^2)^2} = y \frac{x^2-y^2}{x^2+y^2} + xy \frac{4xy^2}{(x^2+y^2)^2} = y \frac{x^2-y^2}{x^2+y^2} + \frac{4x^2y^3}{(x^2+y^2)^2}. \qquad (1)$$

Similarly,

$$f_y (x,y) = x \frac{x^2-y^2}{x^2+y^2} + xy \frac{-2y(x^2+y^2) - 2y (x^2 - y^2)}{(x^2+y^2)^2} = x \frac{x^2-y^2}{x^2+y^2} + xy \frac{-4yx^2}{(x^2+y^2)^2} = x \frac{x^2-y^2}{x^2+y^2} + \frac{-4x^3y^2}{(x^2+y^2)^2}. \qquad (2)$$

Now find partial derivatives at (0,0), using the definition

$$f_x(0,0) = \lim\limits_{\Delta x \to 0} \frac{f(\Delta x, 0) - f(0,0)}{\Delta x} = \lim\limits_{\Delta x \to 0} \frac{0 - 0}{\Delta x} = 0. \qquad (3)$$

$$f_y(0,0) = \lim\limits_{\Delta y \to 0} \frac{f(0, \Delta y) - f(0,0)}{\Delta y} = \lim\limits_{\Delta y\to 0} \frac{0 - 0}{\Delta y} = 0. \qquad (4) $$

So now you can find second derivatives. Again you have to use the definition.

$$f_{xy} (0,0) = \lim\limits_{\Delta y \to 0} \frac{f_x (0, \Delta y) - f_x(0,0)}{\Delta y} \overset{(1),(3)}{=} \lim\limits_{\Delta y \to 0} \frac{ -\Delta y - 0}{\Delta y} = -1.$$

$$f_{yx} (0,0) = \lim\limits_{\Delta x \to 0} \frac{f_y (\Delta x, 0) - f_y(0,0)}{\Delta x} \overset{(2),(4)}{=} \lim\limits_{\Delta x \to 0} \frac{ \Delta x - 0}{\Delta x} = 1.$$