Let $\Phi(u,v)$ differentiable on the plane.
a. which condition should $$\Phi(x+az,y+bz)=0\quad(a,b\in\mathbb R)$$ satisfy in order to define a function $z=z(x,y)$?
b. Prove that $z=z(x,y)$ is a solution for the ode $$a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=-1$$
About A: According to implicit function theorem we can require $$\Phi(x+az(x,y),y+bz(x,y))=0$$ from there I cannot find any constraints on $\Phi$. We also require $$\frac{\partial \Phi}{\partial z}\neq 0$$. but since it's function of $x+az,y+bz$, the derivative $\Phi^\prime$ is function of $a+b$ so we require $a+b\neq0\Leftrightarrow a\neq -b$. In these terms I think $z=z(x,y)$ is defined. I'm not sure about the first condition (seems I miss something)
About B:
If $z=z(x,y)$ so according to the theorem $$\frac{\partial z}{\partial x}=\frac{\frac{\partial \Phi}{\partial x}}{\frac{\partial \Phi}{\partial z}}=\frac{1}{a+b},\frac{\partial z}{\partial x}=\frac{\frac{\partial \Phi}{\partial y}}{\frac{\partial \Phi}{\partial z}}=\frac{1}{a+b}$$ so after substituing I get $$\frac{a+b}{a+b}=1\neq -1$$ Where am I wrong?