Suppose that the function $f:\mathbb{R}^2\to\mathbb{R}$ is continuously differentiable and that there is a positive number $c$ such that $$\frac{\partial f}{\partial y}(x,y) \geq c,\ \ \forall(x,y)\in\mathbb{R}^2$$ Prove that there is a continuously differentiable function $g:\mathbb{R}\to\mathbb{R}$ satisfying $$f(x,g(x)) = 0,\ \ \forall x\in\mathbb{R}$$ and that if $f(x,y)=0$, then $y = g(x)$.
Proof
Because $\frac{\partial f}{\partial y}(x,y) \geq c$, there is one number $y_0$ such that $f(x,y_0) = 0$. Let $x_0 \equiv x$ for symmetry. Then $f(x_0,y_0) = 0$.
Let $g(x) = y_0$ be a constant function for all $x\in\mathbb{R}$. It is therefore continuously differentiable, and satisfies the requirement that $f(x,g(x)) = 0,\ \ \forall x\in\mathbb{R}$.
If we suppose that $f(x,y) = 0$, then $x$ must be equal to $x_0$ and $y = y_0 = g(x)$.
Concern
If $f(x,y) = \sin(x) + y$, then $f$ clearly satisfies the initial criteria for the problem. Hoewever, the zeros of this function are periodic - it is not true that the function is identically zero along a line where $y=0$. Yet when I imagine this function in space, I know that there IS indeed a function which wiggles along space and is equal to zero. I can't quite figure out what gap I have in my understanding, though I think it lies here.
In your proof, you can't take $g$ to be a constant function because $y_0$ depends on $x$.