Given is the curve
$x(p)=\sin(p)$
$y(p)=\cos(p)$
$z(p)=p$
and a vector field $\mathbf{A}(p)$ defined on it as $\mathbf{A}(p)=\left(\begin{array}{c} \cos(p)\\ -\sin(p)\\ 1\\ \end{array}\right)$
Then this vector can be written in many equivalent ways, say, $\mathbf{A}(p)(1)=\left(\begin{array}{c} y(p)\\ -\sqrt{1-y^2(p)}\\ 1\\ \end{array}\right)$, or perhaps $\mathbf{A}(p)(2)=\left(\begin{array}{c} \sqrt{1-x^2(p)}\\ -x(p)\\ 1\\ \end{array}\right)$
I get confused here... I mean, the vector is a function of the coordinates, but they are interdependent via $p$.
Am I still allowed to apply the chain rule in its general form
$\frac{d \mathbf{A}}{dp}=\frac{\partial \mathbf{A}}{\partial x}\frac{dx}{dp}+\frac{\partial \mathbf{A}}{\partial y}\frac{dy}{dp}+\frac{\partial \mathbf{A}}{\partial z}\frac{dz}{dp}$
Is this last expression true? Because this is what happens when I apply it:
$\frac{d \mathbf{A}(1)}{dp}=\frac{\partial \mathbf{A}}{\partial y}\frac{dy}{dp}=\left(\begin{array}{c} -\sin(p)\\ \frac{y \sin(p)}{\sqrt{1-y^2(p)}}\\ 0\\ \end{array}\right)$
Which doesn't match
$\frac{d \mathbf{A}(2)}{dp}=\frac{\partial \mathbf{A}}{\partial x}\frac{dx}{dp}=\left(\begin{array}{c} -\frac{x \cos(p)}{\sqrt{1-x^2(p)}}\\ -\cos(p)\\ 0\\ \end{array}\right)$
$\frac{d\mathbf{A}}{dp}$ should be always the same expression and should not depend on how I chose to write A. But this is not what happens.... The expressions for $\frac{d\mathbf{A}}{dp}$ turn out different.
When dealing with the multivariable chain rule, I find it easiest to keep things straight by dealing with differentials (total derivatives). Let’s distinguish the function $\overline{\mathbf A}:\mathbb R^3\to\mathbb R^3$ from $\mathbf A:\mathbb R\to\mathbb R^3$ and call the parameterization of the curve $\gamma:\mathbb R\to\mathbb R^3$. We then have $\mathbf A=\overline{\mathbf A}\circ\gamma$, so $\operatorname{d}\mathbf A_{p}=\operatorname{d}\overline{\mathbf A}_{\gamma(p)}\operatorname{d}\gamma_{p}$. Here, $\operatorname{d}\overline{\mathbf A}$ is just the Jacobean matrix of $\overline{\mathbf A}$ and the other two differentials are the derivatives with respect to $p$. In terms of partial and ordinary derivatives of these functions, then, the chain rule says that $${d\mathbf A\over dp} = \pmatrix{\partial_1\overline{\mathbf A}_1 & \partial_2\overline{\mathbf A}_1 & \partial_3\overline{\mathbf A}_1 \\ \partial_1\overline{\mathbf A}_2 & \partial_2\overline{\mathbf A}_2 & \partial_3\overline{\mathbf A}_2 \\ \partial_1\overline{\mathbf A}_3 & \partial_2\overline{\mathbf A}_3 & \partial_3\overline{\mathbf A}_3}{d\gamma\over dp} = \pmatrix{\partial_1\overline{\mathbf A}_1{d\gamma_1\over dp} + \partial_2\overline{\mathbf A}_1{d\gamma_2\over dp} + \partial_3\overline{\mathbf A}_1{d\gamma_3\over dp} \\ \partial_1\overline{\mathbf A}_2{d\gamma_1\over dp} + \partial_2\overline{\mathbf A}_2{d\gamma_2\over dp} + \partial_3\overline{\mathbf A}_2{d\gamma_3\over dp} \\ \partial_1\overline{\mathbf A}_3{d\gamma_1\over dp} + \partial_2\overline{\mathbf A}_3{d\gamma_2\over dp} + \partial_3\overline{\mathbf A}_3{d\gamma_3\over dp}}$$ or, using your notation and keeping in mind that the partial derivatives of $\mathbf A$ are themselves vectors, $${d\mathbf A\over dp}={\partial\mathbf A\over\partial x}{dx\over dp} + {\partial\mathbf A\over\partial y}{dy\over dp} + {\partial\mathbf A\over\partial z}{dz\over dp}.$$ For the function in your question, we have ${\partial\mathbf A\over\partial x}=(0,-1,0)^T$, ${\partial\mathbf A\over\partial y}=(1,0,0)^T$ and ${\partial\mathbf A\over\partial z}=(0,0,0)^T$, so ${d\mathbf A\over dp}=(-\sin p,-\cos p,0)^T$, just as you’d get by differentiating it directly.
In your computation, you in fact are getting the same answer both ways. Continue the calculation by substituting for $x$ and $y$. E.g., $$-{y\sin p\over\sqrt{1-y^2}}=-{\cos p\sin p\over\sqrt{1-\cos^2p}}=-{\cos p\sin p\over\sin p}=-\cos p$$ modulo a sign ambiguity that snuck in when you introduced the square root of a squared quantity into the equations.