Generalize Implicit Differentiation to find Tangent Plane For a function $F(x,y,z)$ with $(a,b,c)$ on the level surface $F(x,y,z)=k$, where $F(x,y,z)=k$ defines $z$ implicitly as a function of $x$ and $y$. Using the chain rule, assuming $F_z(a,b,c)\neq0$ show that at the point $(a,b,c)$,
$\frac{\partial z}{\partial x}=g_x(a,b)=-\frac{F_x(a,b,c)}{F_z (a,b,c)}$
I start by using the chain rule to differentiate w.r.t $x$
$F_x=\frac{\partial F}{\partial x} \frac{dx}{dx}+\frac{\partial F}{\partial y} \frac{dy}{dx} + \frac{\partial F}{\partial z} \frac{dz}{dx}$
In the teacher's hint, he stated that $\frac{dy}{dx}$ is equal to zero, which using this I can solve the question, I don't however understand why $\frac{dy}{dx}=0$. I was wondering if someone could explain this.
 A: Get the total differential of F(x,y,z)=k
$$dF(x,y,z)=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz=0$$
Since z is implicitly defined by x and y
$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$
Now replacing dz to dF
$$dF(x,y,z)=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}(\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy)=0$$
Rearranging the terms
$$dF(x,y,z)=\bigg(\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}\bigg)dx+\bigg(\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial y}\bigg)dy=0$$
To make this equation hold we need to satisfy below conditions
$$\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0 \Rightarrow \frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$$
$$\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial y}=0 \Rightarrow \frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$$
The first condition is what you are looking for.
