Vtyjkyuk

I am asked to show, that

$f(x,y)=begincases xyfracx^2-y^2x^2+y^2spacetextfor, (x,y)neq (0,0)\ 0spacetextfor, (x,y)=(0,0)endcases$

is everywhere two times partial differentiable, but it is still $D_1D_2f(0,0)neq D_2D_1 f(0,0)$

But this does not make much sense in my opinion. Since $f(0,0)=0$ hence it should be equal?

I calculated $fracpartial^2 f(x,y)partial xpartial y$ and $fracpartial^2 f(x,y)partial ypartial x$ and I got in both cases the same result (for $(x,y)neq (0,0)$) which is:

$fracx^6+9x^4y^2-9x^2y^4-y^6(x^2+y^2)^3$

Wolframalpha says, that this is correct.

So is there a mistake in the task? Or do I not understand it?

Also, when I want to show, that $f$ is everywhere two times partial differentiable. Is it enough to calculate $fracpartial^2 fpartial^2 x$ and $fracpartial^2 fpartial^2 y$ and not needed to go by the definition, since we know, that this function is differentiable as a function in $mathbbR$?

What do you think?
Thanks in advance.

The limit definition of partial derivative at $(x,y)=(0,0)$:
$$f_x(0,0)=fracpartial fpartial x(0,0)=lim_hto 0 fracf(0+h,0)-f(0,0)h=fracfrac(0+h)cdot 0cdot ((0+h)^2-0^2)(0+h)^2+0^2-0h=0;\
f_y(0,0)=fracpartial fpartial y(0,0)=lim_hto 0 fracf(0,0+h)-f(0,0)h=fracfrac0cdot(0+h)cdot (0^2-(0+h)^2)0^2+(0+h)^2-0h=0.$$
Note that:
$$f_x(x,y)=fracpartial fpartial x=fracpartial fpartial xleft(fracx^3y-xy^3x^2+y^2right)=fracy(x^4+4x^2y^2-y^4)(x^2+y^2)^2;\
f_y(x,y)=fracpartial fpartial y=fracpartial fpartial yleft(fracx^3y-xy^3x^2+y^2right)=fracx^5-4x^3y^2-xy^4(x^2+y^2)^2.\
$$
The partial derivative of partial derivative:
$$fracpartial^2 fpartial xpartial y(0,0)=
lim_hto 0 fracf_x(0,0+h)-f_x(0,0)h=\
lim_hto 0 fracfrac(0+h)(0^4+4cdot 0^2(0+h)^2-(0+h)^4)(0^2+(0+h)^2)^2-0h=lim_hto 0 frac-h^5h^5=-1;\
fracpartial^2 fpartial ypartial x(0,0)=
lim_hto 0 fracf_y(0+h,0)-f_y(0,0)h=\
lim_hto 0 fracfrac(0+h)^5-4(0+h)^3cdot 0^2-(0+h)cdot 0^4((0+h)^2+0^2)^2-0h=lim_hto 0 frach^5h^5=1.$$

First a comment

You wrote But this does not make much sense in my opinion. Since $f(0,0)=0$...

How can you induce an affirmation on second derivatives of a map based on its value at a point?

Second regarding partial second derivatives

You have for $(x,y)neq(0,0)$

$$begincases
fracpartial fpartial x(x,y) = frac(x^2+y^2)(3x^2y-y^3)+2x^2y(y^2-x^2)(x^2+y^2)^2\
fracpartial fpartial y(x,y) = frac(x^2+y^2)(x^3-3x y^2)+2xy^2(y^2-x^2)(x^2+y^2)^2
endcases$$

In particular
$$beginalign*
fracpartial fpartial x(0,y) &= -y text for y neq 0\
fracpartial fpartial x(x,0) &= x text for x neq 0
endalign*$$
which implies
$$
fracpartial^2 fpartial x partial y(0,0) =1 neq -1 = fracpartial^2 fpartial y partial x(0,0)$$

You can have a look here for more details.