$\frac{\partial F}{\partial y}\neq0\implies$ continuous contour line? (Implicit Function Theorem) I have a function $F(x,y)=z$ and two points $(x_1,y_1),(x_2,y_2)$ s.t. $F(x_1,y_1)=F(x_2,y_2)=c$, $x_1<x_2$. I know that $\frac{\partial F}{\partial y}<0$ in $ [x_1,x_2]\times\mathbb{R}$.
I'd like to prove that there is a continuous contour line between the two points.
I know that there's a rectangle $V\times W $ that contains $(x_1,y_1)$ s.t. $F^{-1}(c)\cap V\times W $ is the graphic of a function, i.e., I have in there a continuous contour line.
I'd like to know if the fact that I have $\frac{\partial F}{\partial y}<0$ in the entire interval $[x_1,x_2]$ allows that I consider $V=[x_1,x_2]$ and $W=\mathbb{R}$, so I can prove the statement.
Many thanks!
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Edit for comment on 12/26


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 A: Yes, it's correct. Here's a sketch of the argument. Note first of all that by the implicit function theorem, the portion of a level curve of $F$ in the given region must be a one-dimensional manifold with no boundary points in the open rectangle $(x_1,x_2)\times\Bbb R$.
Let $\Gamma$ be the connected component of the level curve $F(x,y)=c$ passing through $(x_1,y_1)$. The fact that $\partial F/\partial y\ne 0$ tells us that $\Gamma$ can have no vertical tangent line, and this says that the set of $x$-coordinates of points on $\Gamma$ cannot have a maximum value $<x_2$. That is, $\Gamma$ contains a point $(x_2,y^*)$. But now the condition $\partial F/\partial y<0$ tells us that level curves of $F$ can have at most one point for each fixed $x$. Therefore, $y^*=y_2$ and $(x_2,y_2)\in\Gamma$.
A: There may not be a contour line between the two points. Indeed, consider the function
$$F(x,y) = (x^2+1) \arctan y$$
with $\frac{\partial F}{\partial y}(x,y) = \frac{x^2+1}{y^2+1}> 0$ on $\mathbb{R}^2$.
At  the points $(-1,1)$ and $(1,1)$ the functions takes the value $c=\frac{\pi}{2}$, but the level curve $F(x,y) = c$, that is
$$(x^2+1)\arctan y = \frac{\pi}{2}$$
does not intersect the vertical axis $x=0$.
The idea is simple: for every fixed $x$ , the function $F(x,y)$ is strictly increasing in $y$, with image an open interval $I_x$.  As $x$ varies, this interval varies too. It may be  that the interval $I_x$ contains $c$ for $x=x_1$, $x_2$, but, for some intermediate point $x_3$,  $I_{x_3}$ does not contain $c$. The example is chosen so that the $c$ will be one end of the interval $I_{x_3}$.
Note: The level curve $F(x,y) = \frac{\pi}{2}$ is the graph of the function $y = \tan \frac{\pi}{2(x^2+1)}$ defined on $\mathbb{R} \backslash \{0\}$.
$\bf{Added:}$ I changed the sign so that $\frac{\partial F}{\partial y} > 0$, with $F\mapsto -F$.
