How does implicit differentiation relate to multivariable calculus? Assume I don't know much about multivariable calculus (because I don't) I do know about calculus 1 however, and what I'd liked to know is HOW we can go about taking the derivative of an implicit function and WHY we are allowed to do it in the way we do. More specifically, why can we break apart implicit function into an explicit function whose derivatives we can take? I'd very much appreciate visual intuition and rigorous knowledge along with resources someone could point me to.
Thank you.
 A: I'll take a stab at this:
Recall that an implicit function, $f(x,y)$, is just a function that has not yet been solved for one of the variables in terms of the other. That might be because we are not clever enough to solve it, or because it is impossible, or just because we chose not to. It remains an implicit function until it is solved for one of the variables in terms of the other. As an example, consider $x^{2}/16+y^{2}/4=1.$
Suppose we are given a point on the ellipse as $(3,1.3229)$ and asked to find the equation of the tangent line at that point. We know that the derivative of the function evaluated at the given point would give the slope of the tangent line and that using the point slope form of a line we could immediately write the tangent. Can we differentiate the function without solving it?
$$\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$$
Take the derivative of both sides. Of course, the left side is a sum of derivatives, and the right side is zero.
$$D(\frac{x^{2}}{16})+D(\frac{y^{2}}{4})=D(1)$$
When writing these derivatives remember that y is a function of x so write the dy/dx.  That is, we are using the chain rule on the second term, or perhaps we are just using the power rule correctly. $$\frac{1}{16}(2x)+\frac{1}{4}2y\cdot\frac{dy}{dx}=0$$  Next use algebra to solve for $dy/dx$ which is of course the slope of the tangent line.
$$\frac{dy}{dx}=\frac{-x}{4y}=slope=m$$ $$\frac{dy}{dx}=\frac{-x}{4y}=\frac{-3}{4\cdot 1.3329}=-.5669=m$$
Next write the equation using point slope form. $$y-1.3229=m(x-3)$$ and solve for $y$ to write the tangent equation. $$y=-0.5669x+3.0237$$
** Implicit differentiation may also refer to a third and implied variable, such as time. This problem type is often called related rates.  If you google "implicit differentiation and related rates" you should get a few thousand hits with pretty good explanations, and even better if you do the same in youtube.
A: There are a couple of different answers to your question.
The first answer is that the Implicit Function Theorem states that we are fully allowed to treat an implicit function as if it were an explicit function if we can "zoom in" enough on the function so that it looks like an explicit function within a neighborhood.
The reason for this is that an implicit function can always be transformed into an explicit function by adding a parameter which tells you which particular function you are looking at.  Additionally, this parameter can be thought of as a constant within the neighborhood, and therefore doesn't affect differentiation.
Imagine $y = \pm\sqrt{x}$.  $y$ is not a true function of $x$.  However, $y$ generally acts like a function if it is sufficiently zoomed in.  Therefore, we can imagine a parameter $q$ which is attached to the function, which distinguishing which path to take.  So, instead of $y = f(x)$, $y$ is actually $f(x, q)$.  We can define it like this:
$$ f(x, q) = \begin{cases}+\sqrt{x}, \text{ if } q = 0 \\ -\sqrt{x}, \text{ if } q = 1 \end{cases} $$
This is a differentiable function everywhere except the crossover points.  Additionally, we can imagine $q$ to be constant everywhere except the crossover points as well, so it doesn't actually alter the derivative (except, again, at the crossover points).
Therefore, implicit functions can be treated as full functions so long as they look like functions when sufficiently zoomed in.
However, one of the reasons that you are probably tripping up is that you were taught to take full derivatives.  Personally, I find that this confuses things more than it helps.  I find it more helpful to always focus on differentials first, derivatives second.
Example: The derivative of $x^2$ is $2x$, but the differential of $x^2$ is $2x\,dx$.  Because you aren't differentiating with respect to anything, the process doesn't have to care what you are differentiating with respect to.  Additionally, all of the derivative rules are trivially converted to differential rules.  This seems like a trivial difference until you get to more complicated formulas.  
For instance, if I have $x^2 + z^3 + xy = 5$, I can differentiate this into $2x\,dx + 3z^2\,dz + x\,dy + y\,dx = 0$ - that is just a direct application of differential rules.  Now that I have the differential, finding any derivative is merely solving for it.  If I wanted to find $\frac{dy}{dx}$ I can just solve for it:
$$
2x\,dx + 3z^2\,dz + x\,dy + y\,dx = 0 \\
x\,dy = -2x\,dx - 3z^2\,dz - y\,dx \\
\frac{dy}{dx} = -2 - \frac{3z^2}{x}\frac{dz}{dx} - \frac{y}{x}
$$
The way to read this equation is to say that the slope between y and x depends not only on the particular position on the graph, but also on the slope being used between z and x.
I prefer this process because it unifies the differentiation process between explicit derivatives, implicit derivatives, and multivariable total derivatives.
Additionally, you can move from this to partial derivatives just by setting all of the non-participating differentials to 0.  So to find $\frac{\partial y}{\partial x}$ you just set $dz = 0$, which gives:
$$ \frac{\partial y}{\partial x} = -2 - \frac{y}{x} $$
