(I use the f_{x} notation style here 'cause it's easier than ∂f/∂x.)
My calculus book takes it as a given--or gives it as a given, rather--without any sort of proof even in the appendix.
The limit definition of a derivative, has, however, provided me with almost a proof:
f_{xy} = lim_{Δy→0}( (1/Δy)[f_{x}(x,y+Δy,z) - f_{x}(x,y,z)] ) =
lim_{Δy→0}{ (1/Δy)[ lim_{Δx→0}( (1/Δx)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) ] ) - lim_{Δx→0}( (1/Δx)[ f(x+Δx,y,z) - f(x,y,z) ] ) ] } =
lim_{Δy→0}{ (1/Δy)[ lim_{Δx→0}( (1/Δx)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z) ] ) ] },
Because (1/Δy) is independent of Δx, it is constant in the eyes of the delta-x limit item; so, whether (1/Δy) is on the inside or outside of the delta-x limit doesn't matter:
(1) f_{xy} = lim_{Δy→0}{ lim_{Δx→0}( (1/Δx)(1/Δy)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z) ] ) }
By identical reasoning,
(2) f_{yx} = lim_{Δx→0}{ lim_{Δy→0}( (1/Δx)(1/Δy)[ f(x+Δx,y+Δy,z) - f(x,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z) ] ) },
which is virtually identical to (1), and if it can be shown to be true that
lim_{Δx→0}{ lim_{Δy→0}( [an argument] ) } = lim_{Δy→0}{ lim_{Δx→0}( [an argument] ) }
for any and all arguments for which an exception like that annoying thing where they or their limits don't exist isn't present, then f_{xy} must equal f_{xy}.
It seems intuitively reasonable that if you can cram the "limit" parts up against each other then you can switch the two since nothing in either of them is dependent on the changing variable in the other (since Δx, Δy, and zero are all independent of each other). As yet, though, I'm not comfortable--I need to know/see why the result of compositions of limits must be the same regardless of order (for limits with independent sub-arguments).
This is going to involve (ε, δ)-definitions, isn't it?
Also, where does the "lim_{(x,y)→(C,D)}" form come from? Those seem like they may be relevant to my predicament.