TITLE: Find all functions $F(x,y)$ such as $\frac{\sqrt{3}}{2}\frac{\partial{f}}{\partial{x}}+\frac{1}{2}\frac{\partial{f}}{\partial{y}}=0$
QUESTION [10 upvotes]: How to find all possible functions $f(x,y)$ such as:
$$ \frac{\sqrt{3}}{2}f_x+\frac{1}{2}f_y=0$$
(with $f_x = \frac{\partial{f}}{\partial{x}}$ )

Here's everything I tried:
1) I can guess the most simple cases, $f_x=k$ and $f_y=-k\sqrt{3}$ so the following is a solution (but not all possible functions):
$$ f(x,y)= k(x-\sqrt{3}y)+c$$
2) The equation could be writen as:
$$ cos (\frac{\pi}{6})f_x+sin(\frac{\pi}{6})f_y=0$$
But I culdn't see what can I do with that.
3) By the implicit function theorem I could find that:
$$\frac{1}{\sqrt{3}} = - \frac{f_x}{f_y} = \frac{dy}{dx} $$
So $f(x,y)$ has level curves in the form:
$$y=\frac{x}{\sqrt{3}}+c $$
4) I could also find the level curves by noticing that the equation says that the partial derivative of $f(x,y)$ in the direction of $\langle\sqrt{3},1\rangle$ is zero.
5) Or by solving the differential equation:
$$ dx-\sqrt{3}dy=0 $$
But 3), 4) and 5) couldn't help since I don't know how to find $f(x,y)$ based on its level curves.
6) After some trial and error I could find (and prove) that for any $g:R\rightarrow R$ the following function is ok:
$$ f(x,y) = g(x-\sqrt{3}y) $$
I can also see that $\langle\sqrt{3},1\rangle$ is perpendicular to $\langle1,-\sqrt{3}\rangle$, so that's related to the fact that the partial derivative of $f(x,y)$ in the direction of $\langle\sqrt{3},1\rangle$ is zero.

I can see that I'm very close, but I can't find out:

How to prove that all possible functions are of that form?
How could I find that form by calculations?

If there are different ways to find the general form of $f(x,y)$ or to prove it, I would like to read about them all.

REPLY [4 votes]: Let´s use the method of characteristics.
The equation can be converted to:
$$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\cdot \mathrm{grad}f=0$$
That means $f$ is constant along lines which direction is parallel to $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
Hence we can express those lines by:
$$x-\sqrt{3}y=k$$
For each value of $k$, $f$ assumes a different one. Hence $f$ depends on $k$, in other words $f$ is a function of $k$.
Therefore:
$$f(x,y)=g(x-\sqrt{3}y)$$