How to find $F(u,v)$? 
Let $u=f(x,y)$, $v=g(x,y)$, $f,g \in R^2(\Omega)$ and $\operatorname{rank}\left(\dfrac{D(u,v)}{D(x,y)}\right)\equiv 1$. For all $\vec x\in \Omega$, does there always exist a function $F(u,v)$ that $F_u'^2+F_v'^2\neq0$ and $F(f(x,y),g(x,y))\equiv 0$ in a neighborhood $B(\vec x,\delta)$?

I assume $\dfrac{\partial f}{\partial x}\neq 0$, and use the theorem of implicit function to find a function $x=h(u,y)$, and I assume that if such $F$ exist, then let $G(u,y)=F(f(h,y),g(h,y))$ and find that$$\frac{\partial G}{\partial y}=0,\ \frac{\partial G}{\partial u}=F_u'+F_v'\frac{\frac{\partial g}{\partial x}}{\frac{\partial f}{\partial x}},$$but I don't know whether the $F$ exists and I find it hard to construct and $F$ for which $\dfrac{\partial G}{\partial u}$ is always zero. So how to solve this question?
 A: Consider a fixed point $p=(x_0,y_0)\in \Omega$, let $f(p)=u_0$, $\>g(p)=v_0$, and assume $\nabla f(p)\ne0$, $\>\nabla g(p)\ne0$. Both functions $f$ and $g$ then possess a family of level lines in a suitable neighborhood  of $p$, whereby both families cover this neighborhood in a homogeneous way.
The level lines of $f$ can be found as follows: When ${\partial f\over\partial y}(p)\ne0$ we can solve the equation 
$$\bigl(\Psi(x,y,u):=\bigr)\quad f(x,y)-u=0$$
for $y$ in some neighborhood $U$ of $(x_0,y_0,u_0)$: There is a function $(x,u)\mapsto y=\psi(x,u)$ with
$$\psi(x_0,u_0)=y_0,\qquad f\bigl(x,\psi(x,u)\bigr)-u\equiv0\ .$$
Given a fixed value $u$ near $u_0$ the curve
$$\gamma_u:\quad x\mapsto \gamma_u(x):=\bigl(x,\psi(x,u)\bigr)$$
is the level line of $f$ for this value $u$, since $$f\bigl(\gamma_u(x)\bigr)\equiv u\ .\tag{1}$$
It is not necessary to do the same thing for $g$! The condition
$${\rm rank}\left[{\partial(f,g)\over\partial (x,y)}\right]\equiv1$$
means that $$\nabla f(x,y)\parallel\nabla g(x,y)\qquad\forall\>(x,y)\ .\tag{2}$$ From $(1)$ it follows that $\nabla f(\gamma_u(x)\bigr)\cdot\gamma_u'(x)\equiv0$, and $(2)$ now says that $$\nabla g(\gamma_u(x)\bigr)\cdot\gamma_u'(x)\equiv0$$ as well. But this means that $\gamma_u$ is also a level line of the function $g$ that produces the $v$-values! The $v$-value belonging to the chosen $u$ can be computed by applying $g$ to any point of $\gamma_u$. In other words: We have
$$v=F(u):=g\bigl(x,\psi(x,u)\bigr)\ ,$$
whereby the dependence on $x$ on the RHS is only apparent; e.g., you could choose $x:=x_0$.
