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$$ f(x,y)= \begin{cases} \frac{x^3y-y^3x}{x^2+y^2},\ \ (x,y)\ne(0,0)\\ 0,\ \ (x,y)=(0,0) \end{cases} $$ I have to prove that $$ \frac{\partial^2f}{\partial x\partial y}\ne \frac{\partial^2f}{\partial y\partial x} $$ To be honest, I don't really get this task. If I find second partial derivatives at $(x,y)\ne(0,0)$, I'll get total equality because both of them are continuous. If I find either first partial derivative at $(0,0)$, I'll get $0$. It seems to me that, therefore, second partial derivatives at $(0,0)$ will be zeros too. So, what am I doing wrong?

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Okay, let first take the first derivative $f_x$ and $f_y$: $$ f_x(x,y) = \frac{(3x^2y - y^3)(x^2 + y^2) - (x^3y - xy^3)(2x)}{(x^2 + y^2)^2} = \frac{x^4y + 4x^2y^3 - y^5}{(x^2 + y^2)^2} $$ and so you are right that the first derivative at $(0,0)$ is equal to $0$ since $$ f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{\frac{0}{h^2} - 0 }{h} = 0 $$ And similarly, you can also get that $$ f_y(x,y) = \frac{x^5 - 4x^3y^2 -xy^4}{(x^2 + y^2)^2} $$ and so $$f_y(0,0) = \lim_{h \to 0} \frac{f(0,h) - f(0,0)}{h} =0 $$

NOW, THE PROBLEM IS THE SECOND DERIVATIVE: Note the following:

$$ f_{xy}(0,0) = \lim_{h \to 0} \frac{f_x(0,h) - f_x(0,0) }{h} = \frac{(-h^5 -0)/h^4}{h} = -1 $$

whereas

$$ f_{yx}(0,0) = \lim_{h \to 0} \frac{f_y(h,0) - f_y(0,0) }{h} = \frac{(h^5 -0)/h^4}{h} = 1 $$

Therefore, they are not the same!! That is $$ \frac{\partial^2}{\partial x \partial y} \neq \frac{\partial^2}{\partial y \partial x} $$

This may seem strange as you mentioned because it may looks as if it violate the Clairaut's theorem! However, it actually not... because for Clairaut's theorem to be applied, the second derivatives $f_{xy}$ and $f_{yx}$ must be continuous but it's NOT.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Bonrey Jun 6 '19 at 18:34
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Just because the first partial derivative at $(0,0)$ is equal to zero doesn't mean the second partial derivative at $(0,0)$ is equal to zero (think of a parabola at $0$). Finding the values of

\begin{align*} \frac{\partial f}{\partial x}(0,0) && \text{and} && \frac{\partial f}{\partial x}(0,0) \end{align*}

are necessary so that you can then proceed to find the values of the second derivative, but you don't find this second derivative just by taking the partial derivative of these results. Indeed, these are values at a single point: the derivative is a local property.

So, you will need to compute the first partial derivatives away from $0$, and then compute the second partial derivatives at zero (using the limit definition) based on what you find and the values at $(0,0)$ (which, as you pointed out, are both zero).

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