Let $\Phi:\Bbb{R}^2\to \Bbb{R},w=\Phi(u,v)$ a $C^1$ map and let $a,b\in \Bbb{R}$ such that $\Phi(0,0)=0$ and $a\Phi_u(0,0)+b\Phi_v(0,0)\ne 0$. Show there exists a neighborhood of $(0,0,0)\in \Bbb{R}^3$ in which:
a. The equation $\Phi(x-az,y-bz)=0 $ expresses $z$ in terms of $x,y$ such that the expression is a $C^1$.
b. The function (mentioned), $z(x,y)$, satisfies: $a{\partial z\over \partial x}+b{\partial z\over \partial y}=1$.
Right now I am trying to solve (a). I seems a little complicated because it is not in the form that allows to use the theorem. The function should be in three variables. Could I take: $F(x,y,z)=\Phi\circ g(x,y,z)=\Phi(g(x,y,z))=\Phi(x-az,y-bz)=0$? What else is needed to apply the theorem?
As for (b), it was stated that (there are many slightly different formulas for this, as I noticed): $DF_{(x,y)}=-(D(F|_{(x,y)=(0,0)})_{z=0})^{-1}\circ D(F|_{z=0})_{(x,y)=0}$. The notation really confuse me, and I will need some help here as well.
a. You can do just what you said - take $f(x,y,z) = \Phi(x-az, y-bz) = 0$ and differentiate according to z. This gives us $$\frac{\partial f}{\partial z} = D_\Phi(x-az, y-bz)\cdot (-a, -b) |_{(x,y,z)=(0,0,0)}\neq 0$$ Then using the Implicit Function Theorem to get $z = z(x,y)$ as needed.
b. It's almost like you can divide and "reduce" the fractions (just a way to remember! also don't forget the minus) so if $x_i = g(x_j)$ we get $\frac{\partial x_i}{\partial x_j} = - \frac{\frac{\partial g}{\partial x_j}}{\frac{\partial g}{\partial x_i}}$. So we get: $$(\frac{\partial z}{\partial x} , \frac{\partial z}{\partial y}) = -(\frac{\partial f}{\partial z})^{-1} (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$$ And substituting gives us: $$(\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}) = \frac{-1}{-(a\Phi_x + b\Phi_y)} (\Phi_x, \Phi_y)$$ So $a\frac{\partial z}{\partial x} = \frac{a\Phi_x}{a\Phi_x+b\Phi_y}$ and $b\frac{\partial z}{\partial y} = \frac{b\Phi_x}{a\Phi_x+b\Phi_y}$.