so we know that for a function's mixed partial derivatives to be symmetrical we need their second partial derivatives to be continuous. In this example f(x,y) = \begin{array}{l l} \dfrac{xy(x^2-y^2)}{x^2+y^2} & \quad \text{for $(x,y) \neq (0,0)$}\\ 0 & \quad \text{for $(x,y)=(0,0)$} \end{array} they $Fxy$ and $Fyx$ are not the same in $(0,0)$, but I'm not so sure how to prove their second derivatives are not continous in that point, how would I go about it?
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Consider, for example, the one-variable functions $g(x) = f(x,x)$ and $h(x)=f(x,2x)$. – OR. Jun 16 '17 at 11:04
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1Essentially the idea, which might be useful for other problems as well, is to study the restriction of the multivariable function to lines. – OR. Jun 16 '17 at 11:08
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Hint:
We have: $$ \frac{\partial ^2}{\partial x^2}f(x,y)=-\frac{4xy^3(x^2-3y^2)}{(x^2+y^2)^3} $$
now take the limit $(x,y)\to(0,0)$ on the paths $y=x$ and $y=-x$
Emilio Novati
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Yes. This proof that for two different paths the limits are different and the limit for $(x,y)\to(0,0)$ is path-dependent ( so it doe not exists. – Emilio Novati Jun 16 '17 at 12:16