Prove $\nabla u \times \nabla v =0$ is a necessary and sufficient condition that $u$ and $v$ are functionally related by the equation $F(u,v) = 0$ 
Let $u$ and $v$ be differentiable functions of $x$,  $y$ and $z$. Show
  that a necessary and sufficient condition that $u$ and $v$ are
  functionally related by the equation $F(u,v) = 0$ is that $$\nabla u
 \times \nabla v =0$$

 A: Given two functions $u$ and $v$ that are "functionally dependent" the function $F$ exhibiting this dependence is not uniquely determined: The same dependence can  be written, e.g., as $F(u,v):=u^2+v^2-1=0$ or as $G(u,v):=v-\sqrt{1-u^2}=0$. In the following I shall  write the dependence in the form $v=f(u)$ for some function $f$ of type ${\mathbb R}\to{\mathbb R}$. The following theorem is of a purely local nature.
Theorem. Assume that $u$, $v: \Omega\to{\mathbb R}$ are smooth in an open set $\Omega\subset {\mathbb R}^3$, and that $\nabla u(p)\ne0$. Then the following are equivalent:
(a)$\ \quad \nabla u(x)\times\nabla v(x)\equiv0$ in a suitable neighborhood $U$ of $p$,
(b)$\quad$ There is a smooth real-valued function $f$ such that $$v(x)\equiv f\bigl(u(x)\bigr)\tag{1}$$ $\quad\quad\ $ in a suitable neighborhood $U$ of $p$.
Proof. The part (b)$\Rightarrow$(a) is easy: From $v=f\circ u$ we obtain, using the chain rule,
$$\nabla v(x)= f'\bigl(u(x)\bigr)\>\nabla u(x)\qquad\forall x\in U\ .$$
Here $\lambda(x):=f'\bigl(u(x)\bigr)$ is a scalar function. This immediately implies (a).
In order to prove (a)$\Rightarrow$(b) we begin by proving that $v$ is constant on the level surfaces of $u$. Since $\nabla u(p)\ne0$ we may assume that $u_3(x)\ne 0$ in a suitable neighborhood of $p$. The implicit function theorem then allows to solve the equation $$\bigl(H(x_1,x_2,x_3,\bar u):=\bigr)\quad u(x_1,x_2,x_3)-\bar u=0\tag{2}$$ for $x_3$ in a four-dimensional window $W$ with center $\bigl(p_1,p_2,p_3,u(p)\bigr)$: There is a smooth function $\psi$ such that $(2)$ is equivalent with
$$x_3=\psi(x_1,x_2,\bar u)\tag{3}$$
in $W$. For given $\bar u$ near $u(p)$  this equation represents the level set $N_{\bar u}: \ u(x_1,x_2,x_3)=\bar u$ as a graph over the $(x_1,x_2)$-plane, and $(3)$ tells us that $N_{\bar u}$ is indeed a smooth surface.
Consider now a curve $\gamma: \> t\mapsto x(t)\in N_{\bar u}$. Then $\phi(t):=u\bigl(x(t)\bigr)$ has constant value $\bar u$. It follows that $$\nabla u\bigl(x(t)\bigr)\cdot x'(t)=\phi'(t)\equiv0\ .$$
We now look at the value of $v$ along $\gamma$, i.e., consider the function $\chi(t):=v\bigl(x(t)\bigr)$.
Since by our assumption (a) we know that $\nabla v(x)=\lambda(x)\nabla u(x)$ for a certain function $\lambda(\cdot)$ we find that
$$\chi'(t)=\nabla v\bigl(x(t)\bigr)\cdot x'(t)=\lambda(x)\>\nabla u\bigl(x(t)\bigr)\cdot x'(t)\equiv0\ .$$
This shows that $v$ is constant along $\gamma$, whence on all of $N_{\bar u}$.
It follows that the value of $v$ is completely determined by the value $\bar u$ of $u$, so that there is indeed a function $f$ such that $(1)$ holds. It remains to prove that this $f$ is smooth. To this end we invoke $(3)$. For all $(x_1,x_2,\bar u)$ the point $\bigl(x_1,x_2,\psi(x_1,x_2,\bar u)\bigr)$ lies in $N_{\bar u}$. We therefore have
$$f(\bar u)=v\bigl(x_1,x_2,\psi(x_1,x_2,\bar u)\bigr)\ ,$$
where the dependence on $x_1$, $x_2$ on the right hand side is only apparent.
