Why can partial derivatives be exchanged?

There's this little proof, which uses differentiation under the integral sign. $$\begin{align} \frac{\partial f}{\partial x}(x,y) &= \frac{\partial f}{\partial x}(x,y_0) + \int_{y_0}^y \frac{\partial^2f}{\partial x \partial y}(x,t)\,{\rm d}t \\ \frac{\partial^2f}{\partial y \partial x}(x,y) &= \frac{\partial}{\partial y}\int_{y_0}^y \frac{\partial^2f}{\partial x \partial y}(x,t)\,{\rm d}t \\ \frac{\partial^2f}{\partial y \partial x}(x,y) &= \frac{\partial^2f}{\partial x \partial y}(x,y)\end{align}$$

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Another one ("TVM" stands for "mean value theorem"):

Note that this second proof also shows that we only need that both partial derivatives exist, with only one of them being continuous. It is a stronger version.

Here's a nice proof online,but I'm puzzled by the fact you can't find a full proof. Most good multivariable calculus books I've seen, like Bandaxall and Liebeck, have detailed proofs. Anywho, here's a good proof.

https://unapologetic.wordpress.com/2009/10/15/clairauts-theorem/

Basically, the intuitive reason is because the paths in the open set for each variable trace out to the same points in the open set regardless of what the order of partial differentiation is. That is,the tangent lines to the paths are parallel along the mixed partial derivatives curves.

More sophisticated and excellent discussions can be found here.