Let $f(x,y)=xy(x^2-y^2)/(x^2+y^2)$ with f(0,0)=0. Show that $f_1$, $f_2$, $f_{12}$ and $f_{21}$ exist everywhere but $f_{12}$≠$f_{21}$
I tried to solve question by definition of partial derivative but it was very complicated and i think i should try another way to prove that.But i couldn't found. Is there any idea about that?
For $$f(x,y)= \begin{cases} xy\frac{x^2-y^2}{x^2+y^2},&(x,y)\neq(0,0),\\ {0}&(x,y)=(0,0)~. \end{cases} $$ By definition $$f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0$$ $$f_y(0,0)=\lim_{h\to0}\frac{f(0,h)-f(0,0)}{h}=\lim_{h\to0}\frac{0}{h}=0$$ which are exist and for points out of $(0,0)$ we see \begin{eqnarray*} && f_x=\frac{(3x^2y-y^3)(x^2+y^2)-2x(x^3y-xy^3)}{(x^2+y^2)^2} ~~~\text{so with}~~~ y\neq0 ~:~ \color{blue}{f_x(0,y) =-y},\\ && f_y=\frac{(x^3-3xy^2)(x^2+y^2)-2y(x^3y-xy^3)}{(x^2+y^2)^2} ~~~\text{so with}~~~ x\neq0 ~:~ \color{blue}{f_x(x,0) =x}. \end{eqnarray*} and also from $f_x(0,0)=f_y(0,0)=0$ then $$f_{xy}(0,0)=\lim_{h\to0}\frac{f_x(0,h)-f_x(0,0)}{h}=\lim_{h\to0}\frac{-h-0}{h}=-1$$ $$f_{yx}(0,0)=\lim_{h\to0}\frac{f_y(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{h-0}{h}=1$$ this concludes $\color{blue}{f_{xy}(0,0)\neq f_{yx}(0,0)}$.