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Why is it that if $f(x,y)=f(y,x)$ then $\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(y,x)}{\partial y}$ for all $x,y$ in $\Bbb R^2$. My lecturer just went over it like it was obvious but I cant seem to come up with a proof with why it is so. I thought maybe starting from the limit definition of a partial derivative would get me somewhere then I could jumble it till they were equal but this got me nowhere.

Any help would be very much appreciated.

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    Perhaps when we derivate $f(x,y)$ about first coordinate, it's the same which we derivate about first coordinate of $f(y,x)$ – 2017-02-08

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The formula is correct as long as you interpret it as $$ \partial_1 f(x,y) = \partial_2 f(y,x). $$ Indeed, $$ \begin{align} f(x,y) & =f(y,x) \\ f(x+h,y) &= f(y,x+h). \end{align} $$ Now subtract and divide by $h$.

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    Yes, that interpretation question is key here. This formula is a bit of sleight of hand. – 2017-02-08
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Not true at all. For example, try $f(x,y) = xy$.

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    As I read it, the LHS is taken at the point $(x,y)$, the RHS at the point $(y,x)$. – 2017-02-08
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    I do not think you are right... Here $(x,y)$ is the point where we compute the derivative, not the variables... – 2017-02-08