Showing that $f_{xy}(0,0) \neq f_{yx}(0,0)$ for $xy \cdot \frac{x^2-y^2}{x^2+y^2}$

Question

Let the function $\mathbb{R^2} \to \mathbb{R}$ be defined by

$f(x,y) := \begin{cases}xy \cdot \frac{x^2-y^2}{x^2+y^2}, \text{ if } (x,y) \neq (0,0)\\ 0, \text{ else }\end{cases}$

Calculating $f_{xy}(x,y)$ and $f_{yx}$ for $(x,y) \neq (0,0)$ gives

$$f_x(x,y) = \dfrac{y\left(x^2-y^2\right)}{x^2+y^2}+\dfrac{2yx^2}{x^2+y^2}-\dfrac{2yx^2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}$$

$$f_y(x,y) = -\dfrac{2xy^2}{y^2+x^2}+\dfrac{x\left(x^2-y^2\right)}{y^2+x^2}-\dfrac{2xy^2\left(x^2-y^2\right)}{\left(y^2+x^2\right)^2}$$

$$f_{xy}(x,y) = -\dfrac{2y^2}{y^2+x^2}+\dfrac{x^2-y^2}{y^2+x^2}+\dfrac{2x^2}{y^2+x^2}-\dfrac{2y^2\left(x^2-y^2\right)}{\left(y^2+x^2\right)^2}-\dfrac{2x^2\left(x^2-y^2\right)}{\left(y^2+x^2\right)^2}+\dfrac{8x^2y^2\left(x^2-y^2\right)}{\left(y^2+x^2\right)^3} $$

$$f_{yx}(x,y) = \dfrac{x^2-y^2}{x^2+y^2}+\dfrac{2x^2}{x^2+y^2}-\dfrac{2y^2}{x^2+y^2}-\dfrac{2x^2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}-\dfrac{2y^2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}+\dfrac{8y^2x^2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^3}$$

How can one show that $f_{xy}(0,0) \neq f_{yx}(0,0)?$

If I put the values in the functions the denominators would be $0$ and that is undefined.

Answer 1

$$f_{xy}=\lim_{h\to0}\dfrac{f_y(h,0)-f_y(0,0)}{h} = \lim_{h\to0}\dfrac{\lim_{k\to0}\dfrac{f(h,k)-f(h,0)}{k}-\lim_{k\to0}\dfrac{f(0,k)-f(0,0)}{k}}{h}$$

This gives $$f_{xy}=\lim_{h\to0}\dfrac{\lim_{k\to0}\dfrac{hk\frac{h^2-k^2}{h^2+k^2}-0}{k}-0}{h}=\lim_{h\to 0}\frac{h}{h}=1$$

Similarly $$f_{yx}=\lim_{k\to0}\dfrac{f_x(0,k)-f_x(0,0)}{k} = \lim_{k\to0}\dfrac{\lim_{h\to0}\dfrac{f(h,k)-f(0,k)}{h}-\lim_{h\to0}\dfrac{f(h,0)-f(0,0)}{h}}{k}$$ This gives $$f_{xy}=\lim_{k\to0}\dfrac{\lim_{h\to0}\dfrac{hk\frac{h^2-k^2}{h^2+k^2}-0}{h}-0}{k}=\lim_{k\to 0}\frac{-k}{k}=-1$$