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$$f(x,y) = \begin{cases} \displaystyle \frac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y) \neq (0,0), \\ 0 & \text{if } (x,y) = (0,0). \end{cases}$$

I tried finding both mixed partial derivatives but they ended up being the same for that function. I must not be taking into account something dealing with the fact that it is piece-wise. I still need to show the mixed partial derivatives exist. How can I do all of this?

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Use the definition of partial derivative: $$ f_x(0,0) ~=~ \lim_{h\to 0} \frac{f(h,0)-f(0,0)} h ~=~ \lim_{h\to 0} \frac{\frac{h\cdot 0(h^2-0^2)}{h^2+0}-0} h ~=~ \lim_{h\to 0} \frac{0}{h} ~=~ 0 $$ A similar computation shows that $f_y(0,0)=0$ too, so that $(0,0)$ is a critical point (i.e. $\nabla f(0,0)=\binom 00$).


Now that we know the values of $f_x(0,0)$ and $f_y(0,0)$ we can compute $f_{xy}(0,0)$ and $f_{yx}(0,0)$: $$ f_{xy}(0,0) ~=~ \lim_{k\to 0} \frac{f_x(0,k)-f_x(0,0)}k ~=~ \lim_{k\to 0} \frac{f_x(0,k)}k $$ and $$ f_{yx}(0,0) ~=~ \lim_{h\to 0} \frac{f_y(h,0)-f_y(0,0)}h ~=~ \lim_{h\to 0} \frac{f_y(h,0)}h $$ First, note that for $(x,y)\neq(0,0)$ you have $$ f_x(x,y)=\frac{y\big(x^4+4x^2y^2-y^4\big)}{\big(x^2+y^2\big)^2} $$ and $$ f_y(x,y)=\frac{x\big(x^4-2x^2y^2-y^4\big)}{\big(x^2+y^2\big)^2} $$ so that for $h,k\neq 0$ $$ f_x(0,k)=-k \quad\text{and}\quad f_y(h,0)=h $$ Putting all together: $$ f_{xy}(0,0) ~=~ \lim_{k\to 0} \frac{f_x(0,k)}k ~=~ \lim_{k\to 0}\frac{-k}{k} ~=~ -1 $$ and $$ f_{yx}(0,0) ~=~ \lim_{h\to 0} \frac{f_y(h,0)-f_y(0,0)}h ~=~ \lim_{h\to 0} \frac{f_y(h,0)}h ~=~ \lim_{h\to 0}\frac{h}{h} ~=~ 1 $$ so $$ f_{xy}(0,0)~=~-1 ~~\neq~~ 1~=~f_{yx}(0,0) $$

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The mixed partial derivatives are $f_{xy}$ and $f_{yx}$. – TonyK Apr 12 '13 at 16:15
Oops, yeah, I didn't see the `mixed' part... I'm editing right away – AndreasT Apr 13 '13 at 22:15

See here for the hint.

Partial derivative problem

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I figured put how they don't equlk. I simply use the definition of the partial derivative (with the limit as h->0). – Williamm Oct 24 '12 at 0:53

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