Implicit differentiation with partial derivatives?! I really don't have idea if I am correct, but I would like to check if what is making sense to me, is indeed correct.
If I have a surface in $R^3$ that corresponds to the equation below:
$$
z = 6 - x - x² - 2y²
$$

When I'm asked to implicit differentiate that, for example, in terms of $\frac{\partial z}{\partial y}$, why should I treat $x$ as a constant?!?!
The reason for why I should do that, in my head, is because we just want to check what a tiny change in $z$ is going to change our $y$, so it's just like taking a derivative of a curve in a plane that intersects my surface, and that plane is described by a fixed $x$ value. 

Because we're fixing $x$ to get that intersection, we treat it as a constant?! Am I right?
Since:
$$
\frac{\partial z}{\partial y} = -4y
$$
Can assume that $-4y$ is going to be the slope of the tangent line for any point of that intersection?! 
For example, the point $\alpha = (0,1,4)$ is on the plane $x = 0$, and satisfies the equation for the surface. Now, if want to get the tangent line on that point, I just need to get another point that is going to follow the slope of $-4$ because $-4y : y = 1 \Rightarrow -4$, and is on the plane. That point can be, for example $\gamma = (0,2,0)$. Now I can get a director vetor $\vec{t}$, and get a parametric equation for that tangent line:
$$
\vec{t} = \alpha - \gamma = (0,-1,4)
$$
$$
x = 0 + \beta \cdot 0\\
y = 1 + \beta \cdot -1\\
z = 4 + \beta \cdot 4\\
$$

Is everything correct? If it isn't, please try to explain what have I done wrong. Thanks!

By definition;
$$\frac{\partial f}{\partial x}(a,b) = \lim_{h \to 0}\  \frac{f(a+h,b) - f(a,b)}{h}$$
Therefore if you wish to compute $z_x$ then $z = z(x,y)$ and so;
$$\frac{\partial z}{\partial x}(a,b) = \frac{d}{dx} z(x,b) \Bigr|_{x=a}$$
Hence; the partial derivative in the direction of $y$ at $(a,b)$ is the same as differentiating $z(x,b)$ i.e holding $y$ as a constant.  
