Let, $f$ be a function on $\mathbb R^2$ such that $\frac{\partial f}{\partial x}(x,y)=\frac{\partial f}{\partial y}(x,y)$ for all $(x,y)\in \mathbb R^2$. Then which is(/are) correct?
$f(x,y)-f(y,x)=(x-y)\frac{\partial f}{\partial x}(x^*,y^*)+(y-x)\frac{\partial f}{\partial y}(x^*,y^*)$.
$f$ is constant on all lines parallel to the line $x=-y$.
$f(x,y)=0$ for all $(x,y)\in \mathbb R^2$.
$f(x,y)=f(-y,x)$ for all $(x,y)\in \mathbb R^2$.
First we consider an example $f(x,y)=(x+y+5)^2$.
Then the condition $\frac{\partial f}{\partial x}(x,y)=\frac{\partial f}{\partial y}(x,y)$ for all $(x,y)\in \mathbb R^2$ is satisfied.
So, clearly option 3 is FALSE.
Now , $f(-y,x)=(x-y+5)^2\not=f(x,y)$. So , option 4 is also FALSE.
Now equation of lines parallel to the line $x=-y$ is $x+y=c.$ Then , on this $f(x,y)=(5+c)^2=constant$ . So option 2 may be true.
I want to prove option 1 and option 2 , if they are correct. How I can do it ?
For option 2, note any such line can be parameterized as $l(t) = (a+t,b-t),-\infty< t < \infty,$ where $a,b$ are real constants. Consider $f(l(t)).$