Implicit function equation $\frac{\partial z}{\partial x}$ calculation $ xz^2+e^{yz}=0 $
I know this equation:
$ 0=Fx \frac{\partial x}{\partial x} + Fy \frac{\partial y}{\partial x} + Fz \frac{\partial z}{\partial x}$
now I transform it to:
$\frac{\partial z}{\partial x}=\frac{-Fx-Fy\frac{\partial y}{\partial x}}{Fz} $
and I get equation
$\frac{\partial z}{\partial x}=\frac{-Fx}{Fz}=\frac{z^2}{2xz+e^{yz}y} $
Can someone explain why $ Fy\frac{\partial y}{\partial x}=0$ ?
I think when we are taking $\frac{\partial z}{\partial x} $ we are treating y as a constant so Fy=0?
 A: Partial derivative notation is ambiguous. But you've correctly identified the crucial missing piece of information that was intended in your calculation — all of the derivatives are meant to be computed subject to the restriction that $y$ is constant.
It's not $F_y$ that's zero, however; $F_y$ still equals $ze^{yz}$. It's $\frac{\partial y}{\partial x}$ that's zero — it's zero because the intent of the calculation is to take derivatives while holding $y$ constant.
Note that if the intent was to take derivatives while holding $z$ constant, then $\frac{\partial y}{\partial x}$ would not be zero.
One notation I've seen for adding in the missing information is 
$$0=F_x \left.\frac{\partial x}{\partial x}\right|_{y \text{ const}} + F_y \left.\frac{\partial y}{\partial x}\right|_{y \text{ const}}  + F_z \left.\frac{\partial z}{\partial x}\right|_{y \text{ const}} $$
or some abbreviation thereof.

For the sake of precision, I should point out what $F_y$ means too. If we write down the function $F(a,b,c) = a c^2 + e^{bc}$  (so that the given equation is the value of $F(x,y,z)$), $F_y$ is shorthand for $F_2(x, y, z)$, where $F_2$ means to take the derivative of $F_2$ in its second place.
