How to find details about function in $\mathbb{R}^2$ if $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in \mathbb{R}^2$ I am trying assignment questions of real analysis and I was unable to solve this question. In fact, I don't have any idea on how to start.
So, I am looking for help here.

Let $f$  be a function on $\mathbb{R}^2$ such that $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in  \mathbb{R}^2$. Now which of the following options are true:

1 $f$ is constant on all lines parallel to the line $x=-y$.
2 $f(x,y)=0 \quad\forall (x,y) \in \mathbb{R}^2$.
3 $f(x,y)=f(-y,x) \quad\forall (x,y) \in \mathbb{R}^2$.
4 $f(x,y)-f(y,x)=(x-y) \frac {df}{dx}(x*,y*) + (y-x) \frac {df}{dy} (x*,y*) \in \mathbb{R}^2$.
I am unable to get any idea on how to use the hypothesis $\frac{df} {dx}= \frac{df}{dy}$ to deduce or contradict  any of the options.
I studied multivariable calculus from Tom M Apostol in case it is of any help.
Please help me. Kindly just give outlines not complete solution.
 A: If $f_x=f_y$, then for every $c\in \mathbb R$,
$$
\frac{d}{dt}f(t,c-t)=f_x(t,c-t)-f_y(t,c-t)=0.
$$
Hence $f$ is constant on the set
$$
L_c=\{(t,c-t):t\in\mathbb R\},
$$
which represents the parametric expression of the straight line
$$
x+y=c.
$$
Thus $f$ has to be of the form
$$
f(x,y)=g(x+y).
$$
To understand that, observe first that $\mathbb R^2=\bigcup_{c\in\mathbb R}L_c$, and in particular, note that $\,(x,y)\in L_{x+y}$.
As $f$ is constant on every $L_c$, then for every $c\in\mathbb R$, there exists a function $g$, such that
$$
f(x,y)=g(c), \quad \text{for all $(x,y)\in L_c$}.
$$
But $c=x+y$, for all $(x,y)\in L_c$, and hence $\,f(x,y)=g(c)=g(x+y)$.
A: Here is a somewhat more geometric explanation of @Yiorgos or @Ted-Shrifin answers.
$f$ satisfies a first order partial differential equation. This equation says something about a directional derivative, namely that the derivative in the direction $ (-1,1)^t $ is 0. To make things clear, we can change variables so the direction $(-1,1)^t$ is a coordinate direction. There are many ways of doing this, for example:
$ (x,y) = u*(-1,1) + v*(1,0) $
i.e.: $x = -u + v$ and $y = u$.
The inverse transformation is $u = y$, $v=x+y$. Think of $f$ as a function of points, that can be written in terms of different coordinate systems. To avoid confusion, let $f^*$ be $f$ in the new coordinates. Explicitly:
$ f^*(u,v) = f(-u + v, u)$
And so, $f(x,y) = f^*(y, x+y)$
Then:
$ \frac{\partial f^*}{\partial u}(u,v) = -\frac{\partial f}{\partial x}(-u+v,v) + \frac{\partial f}{\partial y}(-u+v,v) = 0$
an equation which is easy to solve: $f^*$ is a function of v only (constant in $u$):
$f^*(u,v) = h(v)$ for some function $h$ (for example $h(v) = f^*(0,v)$ ), and going back to the original coordinates: $f(x,y) = f^*(y,x+y) = h(x+y)$.
So, 1) is true, 2) and 3) are false (they may hold for some solution of the PDE, but not for all solutions), and (assuming (x*,y*) means an intermediate point between (x,y) and (y,x)) 4) is also true (the expressions at both sides of the equation are both 0).
A: The gradient of $f$ is parallel to $\vec i + \vec j$, so level curves are orthogonal to that constant vector. That makes lines $x+y=c$ the level curves, so $f(x,y) = g(x+y)$ for any differentiable function $g$.
A: Since $\displaystyle \frac{\partial f}{\partial x}= \frac{\partial f}{\partial y}$, then we must have $$\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$$ Therefore $$\nabla f= \frac{\partial f}{\partial x}(1,1)$$ and the gradient can be written as $$\nabla f= \frac{\partial f}{\partial x}\vec{i} + \frac{\partial f}{\partial x}\vec{j}$$ This the direction of steepest ascent is along all lines parallel to the vector $\vec{i}+\vec{j}$ Now the gradient is always normal (perpendicular) to the level curves, so this means all lines parallel to $\vec{i}-\vec{j}$ are level curves, that is $f$ is constant along all lines of the forms $x+y= c$, $c\in \mathbb{R}$. Now use that info to make a determination concerning #$3$
