I have exactly the same question. The equation is even more or less the same. Mine is:
f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2} if (x,y) \neq (0,0)
f(x,y)=0 if (x,y)=(0,0).
But I'm still puzzled how to threat the point (0,0). According to my textbook, f_{xy}(0,0)\neq f_{yx}(0,0). So, I know more or less the solution, but I'm not sure that my explanation is correct.
Ignoring the point (0,0) I have:
first x, then y:
\frac{\partial{f}}{\partial{x}}(x,y)=\frac{-y^5+yx^4+4y^3x^2}{(x^2+y^2)^2}
\frac{\partial{f}}{\partial{y}\partial{x}}(x,y)=\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}
first y, then x:
\frac{\partial{f}}{\partial{y}}(x,y)=\frac{x^5-4x^3y^2-y^4x}{(x^2+y^2)^2}
\frac{\partial{f}}{\partial{x}\partial{y}}(x,y)=\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}
So, it is obvious that as long x\neq 0 and y\neq 0 that \frac{\partial{f}}{\partial{y}\partial{x}}(x,y)=\frac{\partial{f}}{\partial{x}\partial{y}}(x,y).
Now, for (x,y)->(0,0), how do I prove that
\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)\neq\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)? Clearly, just substituting x and y with 0 results in \frac{0}{0}.
Is it valid to say the following:
For\frac{\partial{f}}{\partial{y}\partial{x}}(0,0), first let x-> 0 first, resulting in \frac{\partial{f}}{\partial{y}\partial{x}}(0,y)=\frac{-y^6}{y^6}=-1, \forall y and thus\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)=-1?
For\frac{\partial{f}}{\partial{x}\partial{y}}(0,0), first let y-> 0 first, resulting in \frac{\partial{f}}{\partial{x}\partial{y}}(x,0)=\frac{x^6}{x^6}=1, \forall x and thus\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)=1?
This is more or less a guess. I have no clue whether it is correct and if it is correct how to justify this. Especially, I have no explanation why for \frac{\partial{f}}{\partial{y}\partial{x}}(0,0), I first let x->0 and not y->0.
Any help on this would be most welcome and will be highly appreciated.
(and happy new year to you all)
