I was reading a book on mathematical methods for physics by riley hobson and bence and I encountered the following paragraph -
The proof I attempted of that is this - Suppose we want to find the value of $ f(x,y) $ when we increment both x and y by small amounts dx and dy. We can either first increase x, then y or the other way around. The two methods should give the same result : $ \Delta f = (\partial f/\partial x)_{x,y}dx + (\partial f/\partial y)_{x+dx,y}dy = (\partial f/\partial y)_{x,y}dy + (\partial f/\partial x)_{x,y+dy}dx $ Which implies - $ ( (\partial f/\partial x)_{x,y+dy} - (\partial f/\partial x)_{x,y} )dx = ( (\partial f/\partial y)_{x+dx,y} - (\partial f/\partial y)_{x,y} )dy. $ All I glean from this is that the terms next to dx and dy should be zero, and all that says is that the first derivatives are continuous.
The problem with your argument is that you don't take limits anywhere: it isn't true that you can just treat $dx$ and $dy$ as 'small' (it is 'usually' - i.e. when the function is 'nice', but not always).
$$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)_{x,y}=\lim_{dx \to 0}\frac{\frac{\partial f}{\partial y}_{x+dx,y}-\frac{\partial f}{\partial y}_{x,y}}{dx}=\lim_{dx \to 0}\frac{\displaystyle\lim_{dy \to 0}\frac{f_{x+dx,y+dy}-f_{x+dx,y}}{dy}-\displaystyle\lim_{dy \to 0}\frac{f_{x,y+dy}-f_{x,y}}{dy}}{dx}$$
To find the other mixed partial derivative, you need to exchange the order of the limits and this is 'usually' fine, but not in every case: that's why you need the second derivatives continuous.
If you want to follow this through with an example, use $$f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$$ for $x$, $y$ not both equal to $0$ and $f(0,0)=0$. If you evaluate them at $(0,0)$, you should get $1$ for the mixed partial shown above, but $-1$ for the other.