prove $f(x,y) = ax+by$ if and only if $bf_x = af_y$

Question

prove $f(x,y) = ax+by$ if and only if $bf_x = af_y$ also $a,b \neq 0$ and $f$ is $ C^1$ function.

I've got no idea what to do.　I've tried total derivative of $f$ or something like that. I'm also thinking that by being $C^1$ and $bf_x = af_y$ can we conclude that $x$ and $y$ must be separated and both need to have a degree of $1$, but I don't think that seems legit.

Answer 1

The "other way" is not true: $f(x,y)=e^{x+y}$

We have $f_x=f_y$

Answer 2

=> direction - it is trivial

<= direction

See $g(t) = f(c_1 + bt, c_2 - at)$.

$g_t(t) = bf_x(c_1 + bt, c_2 - at) - af_y(c_1 + bt, c_2 - at) = 0$

So, $f(x,y) = h(a*x + b*y)$, where h is $C^1$ function.