Assume f∂x∂y exists and is continuous about the origin. Let m = f∂x∂y(0,0), the mixed partial at 0. Using difference quotients, we will show that f∂y∂x(0,0) also equals m.
As you recall from the previous theorem, there is a "rectangular" function s(h,k) that is shorthand for f(0,0) + f(h,k) - f(h,0) - f(0,k), and s(h,k)/(hk) = f∂x∂y(c,d) for some c and d inside our h by k rectangle. This holds for every h and k near 0.
Let g be the partial of f with respect to y. Write f∂y∂x(0) as a difference quotient in first partials, g(h)-g(0) over h. Then replace g with its limit, giving the following:
f(h,k) - f(h,0) - f(0,k) + f(0,0) over hk
s(h,k) over hk
f∂x∂y(c,d)
Realize that this is a nested limit. We are taking the limit as k approaches 0, then the limit as h approaches 0. But what if we take the limit as h and k approach 0 simultaneously? Since f∂x∂y is continuous, the limit exists, and is equal to m. Now return to the nested limit. Keep h and k sufficiently small, and all the terms that lead to the nested limit are held close to m. The limit as k approaches 0 is close to m, for every h, then the limit as h approaches 0 of the limit as k approaches 0 of s(h,k) is close to m. And "close" is arbitrarily small, within ε. Thus the nested limit has the same value as the 2 dimensional limit. We see that f∂y∂x exists, and is equal to f∂x∂y.