How to think when solving $3\frac{\partial f}{\partial x}+5\frac{\partial f}{\partial y}=0$? Solve this differential equation
$$3\frac{\partial f}{\partial x}+5\frac{\partial f}{\partial y}=0$$
Usually, when we get these problems, they tell us what variable change is smart to do and we just chunk through the chain rule and end up with an answer. Now, you have to think for yourself what variables to put. Therefore, I did not know how to do.
I know the variable substitution that they made, but not why the made it. How should I think here? 
My first guess was $e^{5x-3y}+C$ which does indeed solve it, but that solution is not general enough. Any one function $g(5x-3y)$ will do it, according to the solution manual.

Basically, when given a change of variables $u=u(x,y),v=v(x,y)$, you will use the Chain  rule to transform your equation with unknowns $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ (i.e. the gradient $\nabla f(x,y)$) into an equation with unknowns $\dfrac{\partial f}{\partial u}$ and $ \dfrac{\partial f}{\partial v}$ (i.e. the gradient $\nabla f(u,v)$).
So you may want to think about $0=3\dfrac{\partial f}{\partial x}+5\dfrac{\partial f}{\partial y}$ as $0=\dfrac{\partial x}{\partial u}\dfrac{\partial f}{\partial x}+\dfrac{\partial y}{\partial u}\dfrac{\partial f}{\partial y}=\dfrac{\partial f}{\partial u}$.
So $\dfrac{\partial x}{\partial u}=3$ and $\dfrac{\partial y}{\partial u}=5$, and $x(u,v)=3u+g(v)$ and $y(u,v)=5u+h(v)$. Choos $g$ and $h$ for the transformation to be invertible and computations easy.
For example, take $x=3u-2v$ and $y=5u-3v$ (so $u=-3x+2y$ and $v=-5x+3y$), we obtain that $\dfrac{\partial f}{\partial u}=0$, so $f$ is a constant function of $u$: $f(u,v)=\phi(v)$ for some function $\phi$. Back to the $(x,y)$-coordinates, we have $f(x,y)=\phi(3y-5x)$ for some function $\phi$.
