If q is shorthand for the square of the denominator, the first partials with respect to x and y are as follows.
f∂x = -y5+x4y+4x2y3 over q
f∂y = x5-4x3y2-xy4 over q
Verify that f is 0 on the axes, hence the first partials are 0 at the origin.
Verify that the partials are everywhere continuous. This is clear outside the origin; show that the partials approach 0 as you approach the origin. Restrict x and y to a tiny circle about the origin, of radius r. The denominator is r4, while the numerator of each partial is bounded by a multiple of r5. As r drops to 0, both first partials approach 0.
Verify that the mixed partials exist at the origin. Divide f∂x by y and f∂y by x and take limits. These are the difference quotients, producing the mixed partials at the origin. The mixed partials are -1 and 1 respectively.
We see that mixed partials can be different, even when first partials are everywhere continuous. The problem here is that neither mixed partial is continuous at the origin. Derive the mixed partials yourself and verify this.
If both mixed partials exist near the origin, and are continuous at the origin, they are equal at the origin. This was proved by Bernoulli. (biography)
Choose h and k small, and consider a rectangle with one corner at the origin and the opposite corner at h,k. Let q be a function that adds the value of f at two opposite corners of a rectangle and subtracts the value of f on the other two. For any c between 0 and h:
q(c) = f(0,0) + f(c,k) - f(c,0) - f(0,k)
Since q(c) is a linear combination of values of f, it is differentiable. By the mean value theorem, there is some c where q′(c) is q(h)-q(0) over h. Note that q(0) = 0, hence:
q′(c) = q(h) over h
Let g be the partial of f with respect to x and expand q′, giving:
g(c,k) - g(c,0) = q(h) over h
For any d between 0 and k, let r(d) = g(c,d)-g(c,0). This is also differentiable, since the mixed partials exist throughout the rectangle. Also, the mixed partials are continuous, so apply the mean value theorem again, and for some fixed d:
k×r′(d) = r(k) = q(h) over h
f∂x∂y(c,d) = q(h) over hk
The order of operations isn't important. We could have selected d first, then c. For possibly differen values of c and d, f∂y∂x(c,d) = q(h) over hk.
For all h and k, two mixed partials within the rectangle are equal. As h and k approach 0, mixed partials are equal arbitrarily close to the origin. If mixed partials are continuous at the origin, then the mixed partials are equal at the origin.
If all third partials are continuous throughout an open region, f∂x∂y∂z and f∂x∂z∂y are different second mixed partials of f∂x. Since f∂x satisfies the above criteria for every point in the region, f∂x∂y∂z = f∂x∂z∂y. We can also equate f∂x∂y∂z with f∂y∂x∂z, since f∂x∂y and f∂y∂x are really the same function. If two partials display the same variables, but in a different order, swap adjacent variables until they are equal. This holds for any level of differentiation, provided all the mixed partials remain continuous. We may as well sort variables and group terms together. Hence 7f∂z∂y∂w∂x + 11f∂y∂x∂z∂w = 18 f∂w∂x∂y∂z.