Implicit Differentiation Validity I am interested in a rigorous understanding of how to find the slope of an implicit function, of two variables, at a given point.
So far, I have only found answers as deep as "its an application of the chain rule".
While this may work in practice, I am not convinced this is a rigorous justification by itself, and that there is a deeper justification needed.
Taking the standard definition of differentiation as
$\frac{d}{dx}f(x) = lim_{h \to 0}\frac{f(x+h) -f(x)}{h}$, I find myself unable to apply this directly to implicit functions, given the inability to express the relation as a function, $f(x)$.
I don't doubt the idea of differentiation is very much the same in the implicit, or non-implicit case - but mechanically, given the definition available, I think more work is needed.
My current efforts revolve around seeing the implicit function, $R(x, y) = 0$ as a non-implicit function of two variables, $z = f(x,y)$, then trying to reason about the relation between $\frac{\partial}{\partial x}f(x,y)$ and $\frac{\partial}{\partial y}f(x,y)$ when we constrain $f(x,y) = z_0$. 
Intuition, and a rigorous $\epsilon,\delta$ explanation are both welcomed
 A: Actually it's an case of the implicit function theorem.  In this case the theorem 
says that if $R(x,y)$ is continuously differentiable in a neighbourhood of 
$(x_0, y_0)$, with $R(x_0, y_0) = 0$ and $\frac{\partial R}{\partial y}(x_0, y_0) \ne 0$, then
there is a differentiable function $g$ defined on a neighbourhood of $x_0$
such that $R(x, g(x)) = 0$, and $$g'(x) = - \frac{\partial R/\partial x(x,g(x))}{\partial R/\partial y(x,g(x))}$$
A: It's easier to conceive of when you think of $dy$ and $dx$ as infinitesimal differentials.  Then, they are all just algebraically manipulable just like anything else.  The old Liebnizian method of differentials allows you to differentiate first, and then solve for particular derivatives later.
For instance, if you have the equation $$z = xy + x^2$$ the differential is $$ dz = x\,dy + y\,dx + 2x\,dx $$
This is an equation relating all of the relevant proportions of changes in the variables in the equation.  If you want to know the ratio of two particular differentials (i.e., changes), just algebraically manipulate it so that you have those in ratio.  So, to find $\frac{dy}{dx}$ you just manipulate it algebraically.
$$ \frac{dy}{dx} = \frac{1}{x} \frac{dz}{dx} - \frac{y}{x} - 2 $$
Interpreted, this says that not only does the ratio of the changes in $dy$ and $dx$ depend on $x$ and $y$, it also depends on the ratio of changes between $dz$ and $dx$.
Treating differentials as algebraically manipulable infinitesimals greatly simplifies their treatment across all of calculus.  See "Simplifying and Refactoring Introductory Calculus" for a slightly deeper treatment.
A: For intuition, consider the function $z = f(x,y)$. We can take the total differential of $z$ to be $dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy$. Now since $z$ is constant, $dz = 0$, which gives upon rearrangement that $\frac{dy}{dx} = -\frac{z_x}{z_y}$, which is generally how we use multivariable calculus to justify implicit differentiation.
Also if you learn the implicit function theorem, the result follows as a special case. However the proof behind the theorem is a little involved. 
A: Why do you think the chain-rule explanation is not rigorous? You don't explain this, but I suspect it was the superficial way it might have been presented in, than in any intrinsic fault in thinking of the problem that way. It's perfectly rigorous, as explained below.
The usual problem is this: Given some expression $f(x),$ we want to find its derivative. Although we usually don't, we may write $y=f(x),$ and then we have $y'=f'(x).$ Clearly nothing is affected if this is written as $y-f(x)=F(x,y)=0.$
Now recall that if we have a function of $y,$ say $\phi(y),$ and $y$ in turn is a function of $x,$ say $f(x),$ then the derivative of $y$ with respect to $x$ is given by $\phi'(y)y'.$ But this theorem holds regardless of whether we do actually have an explicit expression $f(x)$ for $y.$ If you see that this is true, then Bob's your uncle.
If you think of any relationship between two variables $$F(x,y)=0$$ as a functional one (with appropriate restrictions), e.g., as $y$ as a function of $x,$ then the function will be well-defined wherever it is, and it matters little that we cannot explicitly write out an expression involving only $x$ for the value $y.$ Thus, as in the usual case we have $y'-f'(x)=0,$ in cases where its impossible to write $y$ explicitly in terms of $x$ alone, we still can differentiate each term of $F(x,y)=0,$ treating each occurrence of $y$ as some unknown expression in $x.$ And this will define the derivative wherever that expression defines a differentiable function.
Thus, given $y^2-f(x)=0,$ we have $$(y^2)'-f'(x)=2yy'-f'(x)=0,$$ where the first term on $\text{LHS}$ of the last equation follows from the chain rule, which is valid for any differentiable function composed with another, regardless of whether an explicit expression is available for it; so long as it is well-defined.
A: This is not an entirely new answer, but meant to complement the existing ones by @Allawonder and @Robert Israel.
As @Allawonder's answer already explained, if by "$y$" we mean nothing more than an alternative notation for a function $f(x)$ we have in mind (e.g., when we are solving a differential equation), then ``implicit differentiation'' holds, simply by the fully rigorous machinery of real analysis: since $y=f(x)$, $F(x,y)$ is just some function $g$ of $x$ which is equal to the constant $0$, and therefore $g'(x)$ must also equal $0$, which gives you an expression for $f'(x)$.
However, the subtlety is that sometimes we are simply given an expression $F(x,y)=0$ without knowing whether such a function $f$ even exists (let alone its derivative). A common example is the equation of the circle $x^2+y^2 - 1 = 0$; at say $x=1$, it simply doesn't even make sense to write $f'(x)$. This is where we need the Implicit Function Theorem, which guarantees us if $F$ is "nice enough" (continuously differentiable), then such a function $y=f(x)$ exists and is differentiable, at least locally; in this case its derivative is given by the usual ``implicit differentiation'' result.
