Some key things to remember about partial derivatives are:
- You need to have a function of one or more variables.
- You need to be very clear about what that function is.
- You can only take partial derivatives of that function with respect to each of the variables it is a function of.
So for your Example 1, $z = xa + x$, if what you mean by this to define $z$
as a function of two variables,
$$z = f(x, a) = xa + x,$$
then $\frac{\partial z}{\partial x} = a + 1$ and
$\frac{dz}{dx} = a + 1 + x\frac{da}{dx},$ as you surmised,
though you could also have gotten that last result by considering $a$ as a
function of $x$ and applying the Chain Rule.
But when we write something like
$y = ax^2 + bx + c,$ and we say explicitly that $a$, $b$, and $c$ are
(possibly arbitrary) constants, $y$ is really only a function of one variable:
$$y = g(x) = ax^2 + bx + c.$$
Sure, you can say that $\frac{\partial y}{\partial x}$ is what happens
when you vary $x$ while holding $a$, $b$, and $c$ constant, but that's
about as meaningful as saying you vary $x$ while holding the number $3$ constant.
I suppose technically $\frac{\partial y}{\partial x}$
is defined even if $y$ is a single-variable function of $x$,
but it would then just be $\frac{dy}{dx}$ (the ordinary derivative),
and I can't remember seeing such a thing ever written as a partial derivative.
It would not make it possible to do anything you cannot do with
the ordinary derivative, and it might confuse people (who might try to
guess what other variables $y$ is a function of).
The previous paragraph implies that the answer to your Example 3 is "yes."
It also hints at why I almost wrote "a function of two or more variables"
as part of the first requirement for using partial derivatives.
Technically I think you only need a function of one or more variables,
but you should want a function of at least two variables before you
think about taking partial derivatives.
For Example 2, where we have $x^2 + y^2 = 1$, it is not obvious
what the function is that we would be taking partial derivatives of.
Either $x$ or $y$ could be a function of the other.
(The function would be defined only over a limited domain,
and would produce only some of the points that satisfy the equation, but
it can still be useful to do some analysis under those conditions.)
If you write something besides the equation to make it clear that
(say) $y$ is a function of $x$, giving a sufficiently clear idea which
of the possible functions of $x$ you mean, then I think technically you
could write $\frac{\partial y}{\partial x}$, and you might even find that
$\frac{\partial y}{\partial x} = 2x$, but again this is a lot of trouble
and confusion to get a result you could get simply by using
ordinary derivatives.
On the other hand, suppose we say that
$$h(x,y) = x^2 + y^2 - 1,$$
and we are interested in the points that satisfy $x^2 + y^2 = 1$,
that is, where $h(x,y) = 0$.
Now we have a function of multiple variables, so we can do interesting
things with partial derivatives,
such as compute $\frac{\partial h}{\partial x}$ and
$\frac{\partial h}{\partial y}$ and perhaps use these to look for trajectories
in the $x,y$ plane along which $h$ is constant.
OK, we don't really need partial derivatives to figure out that
those trajectories will run along circular arcs, but we could have
some other two-variable function where the answer is not so obvious.
In the context of the chain rule, it is understood as you said (and as I said in the comments), in the way that you described as "magically".
Let me give an example to show a slightly different perspective, which hopefully will be the more "natural" interpretation you are looking for.
Consider a circle, with radius $r$. Its area is $A = \pi r^2$. Its circumference is $C = 2\pi r$. We can write $A$ in terms of $C$. While this might seem a little unnatural at first, just bear with me. Take the equation for $C$ and square both sides to get $C^2 = 4\pi^2 r^2$. Now just factor out $4\pi$ to get $C^2 = 4\pi \cdot \pi r^2 = 4\pi A$. Solving for $A$, you get
$$ A = \frac{C^2}{4\pi} $$
Now, we have $A$ is a function of $C$, and $C$ is a function of $r$, so we are in the situation where we could use the chain rule, and it would tell us: $\frac{dA}{dr} = \frac{dA}{dC} \frac{dC}{dr}$. This of course gives $2\pi r$ either way you compute it, but that's not the point. My point is that because of the equation $A = \frac{C^2}{4\pi}$ above, we can make sense of the expression $\frac{dA}{dC}$ without even mentioning $r$! If you just "forget" for a moment that $C$ depends on $r$, and think of $C$ itself as being the independent variable (taking the equation $A = \frac{C^2}{4\pi}$ out of context), then $\frac{dA}{dC} = \frac{C}{2\pi}$ makes perfect sense, as just the ordinary derivative (no magical trick of notation).
My point here was that you can interpret $\frac{df}{dg}$ as just an ordinary derivative, thinking of $g$ itself as the variable (forgetting that $g$ is actually a function). The example above was supposed to illustrate that in many "real-life" situations where the functions have intuitive geometrical meanings, they can be either thought of as variables in their own right, or as functions of another variable. In my circle example, $C$ could either be interprepeted as a function of $r$, or we could just think of $C$ itself as a variable (and ignore the relation between $C$ and $r$).
Best Answer
If you consider $f$ simply as a function of $n$ variables, there is no such definition of "exact" differentiation. The fact is that, sometimes, the variables depend on each other on "external" variables. Let me clarify with an example. Consider a functions $f:\mathbb{R}^2\to\mathbb{R}$ and $u:\mathbb{R}\to \mathbb{R}$. $f$ has two partial derivatives $f_x$ and $f_y$. Suppose that we have a composition $$ \Big( f\circ (\cdot,u(\cdot))\Big)(z) = f(z,u(z)). $$ We can still talk about partial derivatives for $f$, and these are $$ f_x(z,u(z)) \quad \text{and} \quad f_y(z,u(z)). $$ We talk about "total derivative" for $f$ when we differentiate with respect to the parameter to which all the variables depend, namely $z$. This is $$ (f(z,u(z)))' = f_x(z,u(z))+f_y(z,u(z))u'(z). $$ A nice example where both concepts come in action is Euler-Lagrange equation.
The connection with your linked page (differentiable forms) is the following. Suppose you have a function $F:\mathbb{R}^2\to\mathbb{R}^2$. For all $(x,y)\in \mathbb{R}^2$, $F$ is a form of the dual of $\mathbb{R}^2$. The theory developed to understand whether this is also a gradient of a function $f:\mathbb{R}^2\to \mathbb{R}$ requires you to perform integration over curves on the space $\mathbb{R}^2$. Curves are functions $(x(t),y(t))\in \mathbb{R}^2$, $t\in [t_0,t_1]$. If the potential $f$ exists, it's partial derivatives are $$ f_x = F_1 \quad f_y = F_2. $$ On the other side, the derivative of $f$ along a curve $(x(t),y(t))$ is $$ f(x(t),y(t))' = f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t) = F(x(t),y(t))\cdot (x'(t),y'(t)). $$ Therefore you can define the total derivative of $f$ to be $$ df = f_xdx + f_ydy, $$ where $dx$ and $dy$ must be understood as "small" variation of space variables "in time".