If $f(x,y)$ is symmetric (i.e., $f(x,y)=f(y,x)$) and $C^1$ such that its first derivatives $f_x$ and $f_y$ are always strictly positive, is it necessarily the case that there is a $C^1$ function $g(z)$ with strictly positive derivative such that $g(x)+g(y)$ has the same level curves as $f(x,y)$?
It seems to me that this ought to be true, but I am unable to prove it or to find a counterexample. Some sort of smoothness assumption seems to be necessary, as the examples $f(x,y) = \max(x,y)$ and $f(x,y) = x+y+\max(x,y)$ show.
HINT:
Probably not. Such a function should satisfy a certain differential equation.
If a function $h(x,y)$ has the same level curves as $f(x,y)$ then there exists a function $\phi$ smooth bijection so that $$h(x,y)=\phi(f(x,y))$$ The equality $$g(x) + g(y) = \phi(f(x,y))$$ can be replaced with $$g_1(x) + g_2(y) = \phi(f(x,y))$$ due to the symmetry of $f$. Now the LHS are solutions of the equation $\frac{\partial^2 u(x,y)}{\partial x \partial y} =0$. So we are looking for $\phi$ so that $$\frac{ \partial^2}{\partial x\partial y} \cdot (\phi(f(x,y)) )= 0$$ or, equivalently
$$\phi''(f(x,y))\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}+ \phi'(f(x,y))\frac{\partial^2 f(x,y)}{\partial x \partial y} = 0$$
We get $$\frac{\frac{\partial^2 f(x,y)}{\partial x \partial y} }{\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}}=-\frac{\phi''(f(x,y))}{\phi'(f(x,y))}= \psi(f(x,y))$$
Therefore, the gradient of the function $$\frac{\frac{\partial^2 f(x,y)}{\partial x \partial y} }{\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}}$$ is proportional to the gradient of $f(x,y)$. We get a differential equation (PDE) for $f$ of order $3$.
Note: smooth symmetric functions $f(x,y)$ are of form $t(x+y, xy)$. Substituting this, we get a differential equation that $t$ should satisfy.