In my ODE textbook, a problem is written as if $f(x)$ and $g(y)$ have local minima at $x=a$ and $y=b$, respectively, $f(x)+g(y)$ has a minimum at $(a,b)$. Then, there exists a neighborhood $(a,b)$ in which all solutions of the family of the equation $f(x)+g(y)=\text{constant}$ represent closed curves surrounding $(a,b)$.
Can the problem be solved using only $f(x)+g(y)$ has a minimum at $(a,b)$? Are there any necessary conditions such as continuity to prove the above problem?
Let $(a,b)=(0,0)$ and $\psi(x,y):=f(x)+g(y)$. If both $f$ and $g$ have a local minimum at $x=0$, resp., $y=0$ then $\psi$ has a local minimum at $(0,0)$. Conversely: If $\psi$ has a local minimum at $(0,0)$ then $f(x)=\psi(x,0)-g(0)$ has a local minimum at $0$, and $g(y)=\psi(0,y)-f(0)$ has a local minimum at $0$. This was simple.
The thing with the closed curves is less trivial. If both $f$ and $g$ have a nondegenerate local minimum at $0$, i.e., $$f'(0)=g'(0)=0,\quad f''(0)>0, \quad g''(0)>0\ ,$$ then the Morse lemma (applied to $f$ and $g$ separately) guarantees that we can introduce new coordinates $\xi$, $\eta$ in the neighborhood of $(0,0)$ such that $f$ and $g$ appear as $$\hat f(\xi)=f(0)+\xi^2,\quad \hat g(\eta)=g(0)+\eta^2\ .$$ In terms of these new coordinates the level lines of $\hat\psi$ are circles.