For $a$ in the domain, and the function being defined at $a$.
And by $f_1$ I mean the derivative of the function w.r.t to its first argument (so $f_1(x,y)$ means the derivative w.r.t the first argument evaluated at $(x,y)$)
Motivation for the question: i was wondering if there is a relationship b/w $f_1(x,y)$ and $f_1(y,x)$. The answer to this seems to be no, because
if $f(x,y) =(x-y)^2$ then the derivatives are $=$ and of opposite sign
but if $f(x,y) = $(x+y)^2$ then the derivatives are simply equal.
However, for something like $$ \frac{Abs[x-y]}{x+y} $$ the derivatives are not $=$ even in absolute value.
**HOWEVER **All examples I have been able to think of, have the property that the left derivative at $(x,x)$ is equal to the right derivative at $(x,x)$. (as long as the function is defined at $(x,x)$)
Is $f(x,y)$ is symmetric, is there a relationship between $f_1(x,y)$ and $f_1(y,x)$- Do the right and left derivatives need to be equal in absolute value at $(x,x)$ for $f(x,y)$ symmetric -- assuming the function is defined at (x,x)
Edit: I know that we can show that $f_1(x,y) = f_2(y,x)$ with 1 or 2 lines.
I think you're asking whether $|\frac{\partial}{\partial x^-}f(x,y)|_{(x,y)=(a,a)}=|\frac{\partial}{\partial x^+}f(x,y)|_{(x,y)=(a,a)}$ for all $a$ whenever $f$ is symmetric in $x$ and $y$. But that is not true, as seen by $$ f(x,y)=\left\{\begin{array}{lr}x+2y&\text{ if }x\geq y\\2x+y&\text{ if }x<y\end{array}\right. $$ Then $f$ is symmetric with $f_{1^-}(x,x)=2$ but $f_{1^+}(x,x)=1$.