Chain rule: trying to avoid ambiguity in the partial derivatives Given is the curve $\mathbf{\gamma}(p)$ with components
$x(p)=\sin(p)$
$y(p)=\cos(p)$
$z(p)=p$
Then $\frac{d\mathbf{\gamma}(p)}{dp}=\left(\begin{array}{c} \cos(p)\\  -\sin(p)\\  1\\ \end{array}\right)$
Let $\mathbf{A}(p)=\frac{d\mathbf{\gamma}(p)}{dp}$. Expressing $\mathbf{A}(p)$ via the coordinates we can write
$\mathbf{A}(p)(1)=\left(\begin{array}{c} y(p)\\  -\sqrt{1-y^2(p)}\\  1\\ \end{array}\right)$, or
$\mathbf{A}(p)(2)=\left(\begin{array}{c} \sqrt{1-x^2(p)}\\  -x(p)\\  1\\ \end{array}\right)$, or
$\mathbf{A}(p)(3)=\left(\begin{array}{c} y(p)\\  -x(p)\\  1\\ \end{array}\right)$, or
$\mathbf{A}(p)(4)=\left(\begin{array}{c} \cos(z(p))\\  -\sin(z(p))\\  1\\ \end{array}\right)$.
They are different ways to write A.
Now, write the chain rule
$\frac{d\mathbf{A}(p)}{dp}=\frac{\partial \mathbf{A}}{\partial x} \frac{dx}{dp}+\frac{\partial \mathbf{A}}{\partial y} \frac{dy}{dp}+\frac{\partial \mathbf{A}}{\partial z} \frac{dz}{dp}$
Plug in $\mathbf{A}(p)(1)$ and get
$\frac{d\mathbf{A}(p)}{dp}(1)=\left(\begin{array}{c} -\sin(p)\\  \frac{-y \sin(p)}{\sqrt{1-y^2}}\\  0\\ \end{array}\right)=\left(\begin{array}{c} -\sin(p)\\  -\cos(p)\\  0\\ \end{array}\right)$
Plug in $\mathbf{A}(p)(2)$ and get
$\frac{d\mathbf{A}(p)}{dp}(2)= \left(\begin{array}{c} \frac{-x \cos(p)}{\sqrt{1-x^2}}\\  -\cos(p)\\  0\\ \end{array}\right)=\left(\begin{array}{c} -\sin(p)\\  -\cos(p)\\  0\\ \end{array}\right)$
Plug in $\mathbf{A}(p)(3)$ and get
$\frac{d\mathbf{A}(p)}{dp}(3)= \left(\begin{array}{c} 0\\  -1\\  0\\ \end{array}\right)\cos(p)+\left(\begin{array}{c} 1\\ 0\\  0\\ \end{array}\right)(-\sin(p))+0=\left(\begin{array}{c} -\sin(p)\\  -\cos(p)\\  0\\ \end{array}\right)$
Plug in $\mathbf{A}(p)(4)$ and get
$\frac{d\mathbf{A}(p)}{dp}(4)=0+0+ \left(\begin{array}{c} -\sin(z)\\  -\cos(z)\\  0\\ \end{array}\right)=\left(\begin{array}{c} -\sin(p)\\  -\cos(p)\\  0\\ \end{array}\right)$
In all cases the result ends up the same, but the partial derivatives $\frac{\partial\mathbf{A}}{\partial x_i}$ are ambiguous. Still, adding ambiguous terms together consistently results in a correct answer.
User Med suggested that since $y(p)$ is not independent of $x(p)$ etc., we can apply the following chain rule to remove ambiguity:
$\frac{\partial \mathbf{A}}{\partial x_i}= \frac{d\mathbf{A}}{dp}\frac{dp}{dx_i} = \frac{d\mathbf{A}}{dp}\frac{1}{\frac{dx_i}{dp}}$, which works fine in cases (1), (2) and (4), because $\mathbf{A}$ contains only one space variable.
But is case (3) both partial derivatives w/respect to x and y are nonzero and then this workaround can't be used anymore. 
Is there a general approach how to avoid the ambiguity in the partial derivatives? How do I express $\frac{\partial\mathbf{A}}{\partial x_i}$ in case (3)?
 A: I think you're getting tripped up by the notation, which can easily happen when using the derivative notation $\partial \mathbf A/\partial x$.
$\newcommand{\A}{\mathbf A}$
The chain rule says that if $\A(x,y,z)$ is a multivariable function, and then we construct the single-variable function $\A\big(x(p),\ y(p),\ z(p) \big)$ with $x$, $y$, and $z$ functions of another variable $p$, then the derivative of the single-variable function of $p$ is
$$ \frac{d}{dp} \A\big(x(p),\ y(p),\ z(p) \big) = \frac{\partial \A}{\partial x} \frac{dx}{dp}\ +\ \frac{\partial \A}{\partial y} \frac{dy}{dp}\ +\ \frac{\partial \A}{\partial z} \frac{dz}{dp} $$
where the notation $\partial \A/\partial x$ in this context denotes the partial derivative of the multivariable function $\A$ with respect to its first component evaluated at the particular vector $\big(x(p), y(p), z(p) \big)$ for every $p$, and similarly for $\partial \A/\partial y$ and $\partial \A/\partial z$.
The critical thing to note is that $\A$ denotes a multivariable function that takes any triple $(x,y,z)$ and maps it to another triple. But $\A\big(x(p),\ y(p),\ z(p) \big)$ denotes the single-variable function that takes a number $p$ and returns a triple. In other words, it is a composition of a multivariable function with a single-variable function, thus resulting in a single-variable function. That means the notation $\frac{\partial}{\partial x} \A\big(x(p),\ y(p),\ z(p) \big)$ is meaningless: how can we take the partial derivative of a single-variable function?
The difference for your case is that you defined (before you ever mentioned the chain rule) that $\A(p)$ equals something. That makes $\A$ a single-variable function. Thus (strictly speaking) the expression $\partial \A/\partial x$ is meaningless. Although you note that $\A$ can be rewritten so that it looks like a multivariable function. Here's one rewriting (the one you're chiefly interested in):
$$ \mathbf{A}(p)=\left(\begin{array}{c} y(p)\\  -x(p)\\  1\\ \end{array}\right) $$
Even written like this, $\A$ is still merely a single-variable function. But it fits the template of a multivariable function. Namely this one:
$$(x,y,z) \mapsto \begin{pmatrix}
y \\
-x \\
1
\end{pmatrix} $$
$\newcommand{\T}{\mathbf T}$
I'll call this function $\T$ for template. In that case, we may write
$$ \A(p) = \T\big(x(p),\ y(p),\ z(p) \big) $$
and write the chain rule as
\begin{align}
\frac{d\A}{dp} &= \frac{\partial \T}{\partial x} \frac{dx}{dp} + \frac{\partial \T}{\partial y} \frac{dy}{dp} + \frac{\partial \T}{\partial z} \frac{dz}{dp} \\
&= \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} \cos p +
   \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \cdot -\sin p +
   \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \cdot 1 \\
&= \begin{pmatrix} -\sin p \\ -\cos p \\ 0 \end{pmatrix}
\end{align}
as expected.
Of course, it is a lot of trouble to separately define $\T$, and it looks more intuitive to just write $\partial \A/\partial x$. That's fine as long as it's clear from context what your template function is, and you know what's really going on under the hood.
A: This is mainly in response to the answer you provided (and the questions you ask therein). I'd put it as a comment, but it's too long. Hopefully it will help.
$\newcommand{\A}{\mathbf A} \newcommand{\T}{\mathbf T}$
As I mentioned before, you defined $\A$ as a function of $p$. Therefore, strictly speaking, the notation $\partial \A/\partial x$ has no meaning.
So we need to define a meaning for this notation. I take the meaning you're after is "the instantaneous rate of change of the vector $\A$ when $x$ changes by a very small amount". In other words:
$$ \frac{\partial \A}{\partial x} := \lim_{\Delta x \to 0} \frac{\Delta \A}{\Delta x} = \lim_{\Delta p \to 0} \frac{\Delta \A / \Delta p}{\Delta x / \Delta p} = \frac{d\A/dp}{dx/dp} $$
which is the formula you wrote. It should look similar to the formula used to compute the derivative $dy/dx$ when $x$ and $y$ are defined parametrically in terms of a variable $t$.
Now, you wrote $d\A/dp$ using the multivariable chain rule as
$$ \frac{d\A}{dp} = \frac{\partial \A}{\partial x} \frac{dx}{dp} + \frac{\partial \A}{\partial y} \frac{dy}{dp} + \frac{\partial \A}{\partial z} \frac{dz}{dp} $$
But the weird thing is, if we solve the equation
$$ \frac{\partial \A}{\partial x} = \frac{d\A/dp}{dx/dp} $$
for $d\A/dp$, we get
$$ \frac{d\A}{dp} = \frac{\partial \A}{\partial x} \cdot \frac{dx}{dp} $$
without the additional terms! What happened to them?? I think this is where the notational confusion comes in. As I mentioned in my other answer, the multivariable chain rule doesn't really apply to $\A$ since $\A$ is single-variable. However, $\A$ can be made to "look" multivariable if you can match it to a template function $\T$. With the template function (which is multivariable), the chain rule can be properly written as
$$ \frac{d\A}{dp} = \frac{\partial \T}{\partial x} \frac{dx}{dp} + \frac{\partial \T}{\partial y} \frac{dy}{dp} + \frac{\partial \T}{\partial z} \frac{dz}{dp} $$
The thing is, in this answer here, we've given a different definition to $\partial \A/\partial x$, which is incompatible with the template function definition. That is,
$$ \frac{\partial \A}{\partial x} \neq \frac{\partial \T}{\partial x} $$
You can see why if we consider your third case, with the template function
$$ \T(x,y,z) = (y, -x,\ 1) $$
whose partial with respect to $x$ is
$$ \frac{\partial \T}{\partial x} = (0, -1,\ 0) $$
But as for $\partial \A / \partial x$:
\begin{align}
\frac{\partial \A}{\partial x} &= \frac{d\A/dp}{dx/dp} =
\begin{pmatrix}
-\sin p \\
-\cos p \\
0
\end{pmatrix} \cdot \frac{1}{\cos p} \\
&= \begin{pmatrix}
-\tan p \\
-1 \\
0
\end{pmatrix}
\end{align}
which are clearly different for almost all $p$ values.
Interestingly, I think $\partial \A/\partial x$ does equal $\partial \T/\partial x$ in the other cases, but only by a fluke since the template function, though technically dependent on $x$, $y$, and $z$, is really only meaningfully dependent on one of those variables and is constant in terms of the other variables. Thus, for those cases, $\T$ is formally single-variable, and you can get away with confusing the definitions of $\A$ and $\T$ (because the partials with respect to the other variables will be zero, and hence the additional terms of the multivariable chain rule will vanish, but again, I think this is just a fluke of the triviality of the template function).
A: Attempted answer:
This is what I attempted to avoid the ambiguity... Consider unit-speed curves. 
The curve
$x(p)=\sin(p)$
$y(p)=\cos(p)$
$z(p)=p$
is unit-speed and $\mathbf{A}(p)$ is its tangent vector with norm 1.
Above, $\mathbf{A}(p)$ was written in four equivalent ways. But (3) is incorrect, because its components don't contain  the 
relationship they have through $p$ (notice that $\mathbf{A}_x(p),\mathbf{A}_y(p),\mathbf{A}_z(p)$ are interdependent via $p$). The vector (3)
$\left(\begin{array}{c} y\\  -x\\  1\\ \end{array}\right)$
can be considered incorrect, because its norm is not equal to 1, unless we substitute in $x(p)$ and $y(p)$ their definitions. All other 
cases (1), (2) and (4) are fine, because their components show how they are inter-related via $p$ and their norm is 1, thus 
showing that the tangent vector they represent belongs to a unit-speed curve.
Then, let $\mathbf{A}(p)$ be written for now to depend on only one of the spatial variables $x$, $y$ or $z$. This way we arrive at the expression (as per user Med's suggestion)
$\frac{\partial \mathbf{A}}{\partial x_i}= \frac{d \mathbf{A}}{dx_i}=\frac{d \mathbf{A}}{dp} \frac{dp}{dx_i}$
which yields the non-ambiguous formula for the spatial derivatives of $\mathbf{A}(p)$:
$\frac{\partial \mathbf{A}}{\partial x_i}=\frac{\mathbf{A}^{\prime}}{x_i^{\prime}}$
(prime means d/dp). This formula always produces the same result no matter which spatial variable we chose to express $\mathbf{A}$ into!!! One needs to remember, that when we take $\frac{\partial}{\partial x}$ of say, (1), $y$ depends on $x$ via $p$ and thus the partial derivative is nonzero.
Notice that $\frac{\partial (1)}{\partial x}=\frac{\partial (2)}{\partial x}=\frac{\partial (3)}{\partial x}=\frac{\partial (4)}{\partial x}$
It even works on (3) too.
Notice that only one derivative $\frac{\partial \mathbf{A}}{\partial x_i}$ should be treated as nonzero, all others $\frac{\partial \mathbf{A}}{\partial x_j}$ (where $j\ne i$) should be treated as equal to zero. That is, the notation should be consistent. If we chose to express  $\mathbf{A}$ via $x$, we should stay with this notation. Because in this way we ensure that the chain rule is meaningful:
$\frac{d\mathbf{A}}{dp}=\frac{\partial \mathbf{A}}{\partial x}\frac{dx}{dp}+\frac{\partial \mathbf{A}}{\partial y}\frac{dy}{dp}+\frac{\partial \mathbf{A}}{\partial z}\frac{dz}{dp} = \frac{\mathbf{A}^{\prime}}{x^{\prime}}x^{\prime}+0 +0= \mathbf{A}^{\prime}=\frac{d\mathbf{A}}{dp}$
