Why aren't these two limits (calculated with CAS) equal? The first limit is
$$\lim_{(x,y) \to (0,0)} \frac{\partial^2}{\partial y \partial x} \frac{x^3 y -x y^3}{x^2 +y^2},$$
computed by
(%i17)  limit(diff(diff((x^3*y-x*y^3)/(x^2+y^2),x,1),y,1),(y,x),(0,0));
(%o17)  -1

The second limit is
$$\lim_{(y,x) \to (0,0)} \frac{\partial^2}{\partial x \partial y} \frac{x^3 y -x y^3}{x^2 +y^2},$$
computed by
(%i22)  limit(diff(diff((x^3*y-x*y^3)/(x^2+y^2),y,1),x,1),(x,y),(0,0));
(%o22)  1

The CAS is Maxima.
Why do the two outputs differ? What does it result from?

In addition to the judicious answer from Christian Blatter :
$$f(x,y)=\frac{\partial^2}{\partial x\partial y}\left(\frac{x^3y-xy^3}{ x^2+y^2}\right)=\frac{(x^2-y^2)(x^4+10 x^2 y^2+y^4)}{ (x^2+y^2)^3}\qquad
(x,y)\ne(0,0)$$
we observe that :
Case 1 :
$$f(x,0)=1$$
Thus if $y$ is set to $0$ and $x\to 0$ the limit is $1$.
Case 2 :
 $$f(0,y)=-1$$
Thus if $x$ is set to $0$ and $y\to 0$ the limit is $-1$. 
Even more, if  $\quad y=cx\quad$ with $\quad c\ne 0\quad$ we get:
$$f(x,cx)=\frac{(1-c^2)(1+10c^2+c^4)}{(1+c^2)^3}$$
Thus if $x\to 0$ following the path $y=cx$, the limit is $\frac{(1-c^2)(1+10c^2+c^4)}{(1+c^2)^3}$
So, they are as many different limits as many manner to make $x\to 0$ and $y\to 0$ according to various paths relating $x$ and $y$.
In other words, there is no determined limit.
