Why normal vector's formula for implicit function is different than for explicit function? Find tangent plane to surface: $z = x^2 + y^2 - 3$ at point $P(2, 1, 2)$
The normal vector to the plane I am looking for is defined as:
$$\vec{n} = [\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1] \Big|_{P(2,1,2)}$$
I think I understand everything. Problem is when I am dealing with an implicit form of function.
Find tangent plane to $4x^2 + 2y^2 + z^2 - 12 = 0$ at point $P(1, -\sqrt{2}, -2)$
So the normal vector is defined as:
$$\vec{n} = [\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}] \Big|_{P(1, -\sqrt{2}, -2)}$$
I am still getting good results. But I simply cannot understand what is going on. 
The second, implicit function, could be rewritten in explicit form as something like: $$z = \pm \sqrt{-4x^2 - 2y^2 + 12} = f(x, y)$$
So... it is really confusing to me why in the first normal vector's equation there's $-1$ and in the second there's $\frac{\partial f}{\partial z}$ instead. After all I am dealing with function of two variables in both cases, right?
And, after I convert the function from implicit to explicit form (and even before conversion), shouldn't I use $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z}$ notation instead of $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$?
To show what I am struggling with to understand: let's say I have function $y = x^2 = f(x)$. Clearly I can find $y'$, but I cannot find $x'$, right?
Can someone help me understand this, how is it possible that the first normal vector equation can be that much different than the implicit one? 
Are those two functions: implicit and explicit, even equivalent or not?
 A: Note that:$$z=x^2+y^2-3\iff\overbrace{x^2+y^2-z}^{\phantom{f(x,y,z)}=f(x,y,z)}=3.$$So, if you apply both methods, the you get the same results. That $-1$ comes from $\frac{\partial f}{\partial z}$.
A: 
After all I am dealing with function of two variables in both cases, right?

No.  In the second case you're dealing with a function of three variables.  You can see that because you have all three variables on one side of the equation
$$
4x^2 + 2y^2 + z^2 - 12 = 0
$$
and a constant on the other.  Also, as you point out, the normal vector includes $\frac{\partial f}{\partial z}$ as a component, so $f$ must be a function of $z$ (in addition to $x$ and $y$).
If you want to solve the second question using the first method, you can write
$$
    z = - \sqrt{-4x^2 - 2y^2 + 12}
$$
(Not $\pm$, you have to choose the negative branch to pass through $(1, -\sqrt{2}, -2)$.
Then you get
\begin{align*}
    \frac{\partial z}{\partial x}
    &= \frac{1}{2}(-4x^2-2y^2+12)^{-1/2}(-8x) 
     = -\frac{4x}{\sqrt{-4x^2 - 2y^2 + 12}} \\
    \frac{\partial z}{\partial y} 
    &= \frac{1}{2}(-4x^2-2y^2+12)^{-1/2}(-4x)
     = -\frac{2y}{\sqrt{-4x^2 - 2y^2 + 12}} \\
\end{align*}
At $x=1$ and $y=-\sqrt{2}$ this gives a normal vector of
$$
    \vec n_1 = \left[-\frac{4\cdot 1}{-2},-\frac{2\cdot -\sqrt{2}}{-2},-1\right]
             = \left[2,-\sqrt{2},-1\right]
$$
This seems different from the normal vector given by the second method:
$$
    \vec n_2 = \left.\left[8 x,4y,2z\right]\right|_{(1,-\sqrt{2},-2)}    
             = \left[8,-4\sqrt{2},-4\right]
$$
But as you can see $\vec n_2 = 4 \vec n_1$.  “Normal” is more of a line than a vector; any two normal vectors will be linearly dependent.  
