Use Implicit Function Theorem to show that J(x,y)=0 Let $u=f(x,y)$ and $v=g(x,y)$ be defined in a region "R". Suppose there exists another non constant function $h$, of class $C^{1}$, s.t $h(u,v)=0$ $\forall (x,y) \in$ R. Prove that 
$J=\frac{\partial(u,v)}{\partial(x,y)}=0 $, $\forall(x,y)\in$ R. 
What I've managed to prove so far is the following. If you derivate $h(u,v)$ with respect to $x$ and $y$ you get a homogeneous system of linear equations. If this system has non trivial solutions, then the problem is solved. 
I appreciate any help, Thanks!
 A: Derivating $h(u,v)=0$ respect to $x$ and $y$ we have:
\begin{equation}
\frac{\partial{h}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{h}}{\partial{v}}\frac{\partial{v}}{\partial{x}}=0 
\end{equation}
and
\begin{equation}
\frac{\partial{h}}{\partial{u}}\frac{\partial{u}}{\partial{y}} + \frac{\partial{h}}{\partial{v}}\frac{\partial{v}}{\partial{y}}=0
\end{equation}
If we denote $A$ to the system above, then its straightforward to see that
\begin{equation}
det(A)= \frac{\partial{(u,v)}}{\partial{(x,y)}}
\end{equation} 
So , it would be sufficient to prove that $det(A)=0$ $\forall (x,y) \in R$. This will guarantee that the Jacobian is exactly $0$ too. Note that:


*

*If A has only the trivial solution, then (in $R$):
$$\frac{\partial{h}}{\partial{u}}=\frac{\partial{h}}{\partial{v}} = 0$$


This is not possible because, according to the hypothesis, $h$ is a non constant function. So at least one of these derivatives is not $0$. 
$\therefore$ $A$ has a non trivial solution
$\therefore$ $det(A)=0$
\begin{equation}
\therefore
J=\frac{\partial{(u,v)}}{\partial{(x,y)}}=0 
\end{equation}
$\forall (x,y) \in R$  
A: $$
0=\frac{\partial h(u(x,y),v(x,y))}{\partial x}=h_u u_x+h_v v_x,\\
0=\frac{\partial h(u(x,y),v(x,y))}{\partial y}=h_u u_y+h_v v_y,
$$
and hence
$$
(h_u,h_v)\cdot (u_x,v_x)=(h_u,h_v)\cdot (u_y,u_y)=0,
$$
and as $h$ is non-constant, then $(h_u,h_v)$ does not vanish identically, and thus the vectors
$$
(u_x,u_y)\quad \text{and}\quad  (v_x,v_y),
$$
are parallel, and thus 
$$
\left|\begin{array}{cc} u_x & u_y \\ v_x & v_y
\end{array}
\right|=0.
$$
