Gradient/partial derivatives and proof?

Suppose f:R^2->R is a differentiable function whose gradient is nowhere 0 and that satisfies ∂f/∂x = 2(∂f/∂y). Find, with proof, the level curves of f. Show that there is a differentiable function F:R->R s.t. f(x,y) = F(2x+y) So I integrated both sides of the differential equation ∫(∂f/∂x)dx =... show more Suppose f:R^2->R is a differentiable function whose gradient is nowhere 0 and that satisfies ∂f/∂x = 2(∂f/∂y).
Find, with proof, the level curves of f.
Show that there is a differentiable function F:R->R s.t. f(x,y) = F(2x+y)

So I integrated both sides of the differential equation
∫(∂f/∂x)dx = 2∫(∂f/∂y)dy
f(x) = 2f(y)

Somehow, I assume this gets me f(x,y) = 2x + y, since that's used in the second part of the problem. How do I make this conclusion (ie, how do I go from f(x) = 2f(y) to f(x,y) = 2x + y)? It seems like this should be a basic calculus thing, but I keep getting hung up on it... (maybe this is the universe's way of telling me that I shouldn't do math late at night, haha)

If it helps, the topic of this section is supposed to be the gradient and using the gradient to define tangent planes/surfaces (specifically, ∇f(ā) · (x̄-ā) = 0 )
Update: I also have a sketch of the proof:

Suppose g(t) = f(t, -2t)
g'(t) = ∂f/∂x - 2(∂f/∂y) = 0
So g is constant; ie, g has the same value at any point t.

But then I get stuck again--how does this get me to 2x y?
Update 2: The last thing was supposed to say "how does this get me to 2x y?"
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