Multivariable Calculus Help Prove that if the function z is defined implicitly by
$$F( x - az, y -bz ) = 0$$ 
where F has continuous partial derivatives, then 
$$a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y} = 1$$
I have a hunch that I must somehow use the Implicit Function Theorem, but I am not exactly sure how to apply it in this case. Any suggestions?
 A: Differentiating both sides of the equality $$F( x - az,\, y -bz ) = 0 \tag{1}$$ by $x,$ we have
$$\dfrac{\partial}{\partial{x}}\left( {F( x - az, y -bz )} \right)= 0, \\
\left.\dfrac{\partial{F}}{\partial{t}}\right\vert_{t=x-az(x,\,y)}\cdot \dfrac{\partial{}}{\partial{x}}(x-az(x,\,y))+\left.\dfrac{\partial{F}}{\partial{s}}\right\vert_{s=y-bz(x,\,y)}\cdot \dfrac{\partial{}}{\partial{x}}(y-bz(x,\,y))=0.$$
But
$$\dfrac{\partial{}}{\partial{x}}(x-az(x,\,y))=1-a\dfrac{\partial{z}}{\partial{x}},\\
\dfrac{\partial{}}{\partial{x}}(y-bz(x,\,y))=-b\dfrac{\partial{z}}{\partial{x}}.$$
Hence
$$\left(1-a\dfrac{\partial{z}}{\partial{x}}\right)\dfrac{\partial{F}}{\partial{t}}-b\dfrac{\partial{z}}{\partial{x}}\dfrac{\partial{F}}{\partial{s}}=0 .\tag{2}$$
In the same manner, differentiating $(1)$ by $y,$ we obtain
$$-a\dfrac{\partial{z}}{\partial{y}}\dfrac{\partial{F}}{\partial{t}}+\left(1-b\dfrac{\partial{z}}{\partial{y}}\right)\dfrac{\partial{F}}{\partial{s}}=0. \tag{3}$$
Now you can solve the system $(2),\ (3) .$
A: $DF = (D_{1}F\cdot (1 - a\cdot \partial z/\partial x)+D_{2}F\cdot (-b\cdot \partial z/\partial x), D_{1}F\cdot (-a\cdot \partial z/\partial y) + D_{2}F\cdot (1 - b\cdot \partial z/\partial y)) \equiv 0$
So, this reduces to:
$$\begin{pmatrix} 1 - a \frac{\partial z}{\partial x} & -b\cdot \frac{\partial z}{\partial x}\\ 
-a\cdot \frac{\partial z}{\partial y} & 1 - b \frac{\partial z}{\partial y}
\end{pmatrix} \cdot \begin{pmatrix} D_{1}F \\ D_{2}F \end{pmatrix} = 0.$$
Now we want the matrix to be non-singular; otherwise $D_{1}F$ and $D_{2}F$ must both be identically zero. Why is this a problem?
Then the determinant is $1 - a\cdot \frac{\partial z}{\partial x} - b\cdot \frac{\partial z}{\partial y} + (ab-ab)\cdot \frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y} \equiv 0$. Hence, $a\cdot \frac{\partial z}{\partial x} + b\cdot \frac{\partial z}{\partial y} = 1$.
