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)?