Define a function $f$ on $\mathbb{R}^2$ by
$$ f(x,y)=\begin{cases} x^2 \arctan \frac{y}{x} - y^2 \arctan \frac{x}{y} &\text{for}~~~ xy= 0 \\ 0 & \text{otherwise} \end{cases}$$
Show that $f_{xy}(0, 0)$ and $f_{yx}(0, 0)$ exist. Then calculate them to show that $f_{xy}(0, 0) \neq f_{yx}(0, 0)$
How can i approach this question? I thought $f_{xy}(0, 0) = f_{yx}(0, 0)$?
On $U = \{ (x,y) \in \mathbb R^2 \mid xy \ne 0\}$ the function $f$ is clearly differentiable with continuous partial derivatives $$f_x(x,y) = 2x\arctan\frac{y}{x} + x^2\frac{1}{1 + (\frac{y}{x})^2}(-\frac{y}{x^2}) -y^2\frac{1}{1 + (\frac{x}{y})^2}\frac{1}{y} = 2x\arctan\frac{y}{x} - \frac{y}{1 + (\frac{y}{x})^2} - \frac{y}{1 + (\frac{x}{y})^2}$$ and $$f_y(x,y) = \frac{x}{1 + (\frac{x}{y})^2} + \frac{x}{1 + (\frac{y}{x})^2} -2y\arctan\frac{x}{y} .$$ In $(0,y)$ with $y \ne 0$ we get $$f_x(0,y) = \lim_{h\to 0}\frac{f(0+h,y)- f(0,y)}{h} = \lim_{h\to 0}\frac{h^2\arctan\frac{y}{h} - y^2\arctan\frac{h}{y} - 0}{h} \\= \lim_{h\to 0}(h\arctan\frac{y}{h}) - y \lim_{h\to 0}\frac{\arctan\frac{h}{y}-\arctan0}{\frac{h}{y}} = 0 - y\arctan'(0) = -y$$ since $\arctan$ is bounded and $\arctan 0 = 0, \arctan'(0) = 1$.
In $(x,0)$ with arbitrary $x$ we get $$f_x(x,0) = \lim_{h\to 0}\frac{f(x+h,0) - f(x,0)}{h} = \lim_{h\to 0}\frac{0- 0}{h} = 0 .$$ Similarly in $(x,0)$ with $x \ne 0$ we get $$f_y(x,0) = x$$ and for $(0,y)$ with arbitrary $y$ we get $$f_y(0,y) = 0 .$$ Thus the partial derivatives $f_x, f_y$ exist on all of $\mathbb R^2$. We get $$f_{xy}(0,0) = \lim_{h\to 0}\frac{f_x(0,h)- f_x(0,0)}{h} = \lim_{h\to 0}\frac{-h}{h} = -1 ,$$ $$f_{yx}(0,0) = \lim_{h\to 0}\frac{f_y(h,0)- f_x(0,0)}{h} = \lim_{h\to 0}\frac{h}{h} = 1 .$$