Is curl of a particle's velocity zero? The question
Consider the motion of a particle specified by $\mathbf{x} (t): \mathbb{R} \mapsto \mathbb{R}^3$, where $\mathbf{x} = (x_1,x_2,x_3)$ in cartesian coordinates. The curl of its velocity $\mathbf{v} = (v_1, v_2, v_3)$ can be calculated as
$$
\nabla \times \mathbf{v} = (\frac{\partial v_3}{\partial x_2} -\frac{\partial v_2}{\partial x_3},\frac{\partial v_1}{\partial x_3} -\frac{\partial v_3}{\partial x_1}, \frac{\partial v_2}{\partial x_1} -\frac{\partial v_1}{\partial x_2} ).
$$
From the interchangeability of ordinary and partial derivatives,
$$
\frac{\partial v_i}{\partial x_j} 
= \frac{\partial}{\partial x_j} \frac{\mathrm{d}x_i}{\mathrm{d}t} 
= \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial x_i}{\partial x_j} 
= \frac{\mathrm{d}}{\mathrm{d}t} \delta_{ij}
=0,
$$
which makes every component of the curl zero for any $\mathbf{x}(t)$. However, this does not make much sense to me, since I can easily imagine a rotating velocity field. Also, such fields seem to be quite common in rotational dynamics and fluid dynamics, such as in this post.
Some contexts
I am trying to derive the equation of motion for a charged particle with rest charge $m$ and charge $q$ in a given electric potential $\phi(t,\mathbf{x})$ and magnetic vector potential $\mathbf{A}(t,\mathbf{x})$, whose Lagrangian is given by
$$
\mathcal{L} = -mc^2 \sqrt{1-|\mathbf{\dot{x}}|^2/c^2} - q \phi + q \mathbf{\dot{x}} \cdot \mathbf{A}.
$$
The above question arises when evaluating the term $\nabla (\mathbf{\dot{x}} \cdot \mathbf{A})$ (the spatial gradient), which appears in ${\partial \mathcal{L}}/{\partial x_i}$. Using the vector calculus identities, we can evaluate the spatial gradient as
$$
\nabla(\mathbf{A} \cdot \mathbf{\mathbf{\dot{x}}})  =\  (\mathbf{A} \cdot \nabla)\mathbf{\mathbf{\dot{x}}} \,+\,  (\mathbf{\mathbf{\dot{x}}} \cdot \nabla)\mathbf{A} \,+\,  \mathbf{A} {\times} (\nabla {\times} \mathbf{\mathbf{\dot{x}}}) \,+\,  \mathbf{\mathbf{\dot{x}}} {\times} (\nabla {\times} \mathbf{A}) \\
= \,  (\mathbf{\mathbf{\dot{x}}} \cdot \nabla)\mathbf{A} +  \mathbf{\mathbf{\dot{x}}} {\times} (\nabla {\times} \mathbf{A}),
$$
where the second step uses the identity in the question. The answer to a similar question by Shuhao Cao (non-relativistic, but the terms of interest are identical) explains the second step as "the usual postulate that $\mathbf{v}$ is not explicitly a function of $\mathbf{x}$."
 A: $
\newcommand\DD[2]{\frac{d#1}{d#2}}
\newcommand\PD[2]{\frac{\partial#1}{\partial#2}}
\newcommand\DDf[2]{d#1/d#2}
\newcommand\PDf[2]{\partial#1/\partial#2}
\newcommand\R{\mathbb R}
$
We need to be careful with partial derivatives.
If I have a function $f(x(t), y) = x + y$ with $x(t) = t$, then what is $\PDf fx$? On one hand you may say $1$, but on the other hand $f(x(t), y) = t + y$ so why shouldn't $\PDf fx = 0$? You have to be clear about what is varying and what is being held constant. In this case, we could say what we actually have is a function $f_2 : \R^2 \to \R$ defined by $f_2(x, y) = x + y$, another function $X : \R \to \R$ defined by $X(t) = t$, and another function $f_2 : \R^2 \to \R$ defined by $f_2(t, y) = X(t) + y$. Then
$$
  \PD{f_1}x(x, y) = 1,\quad \DD Xt(t) = 1,\quad \PD{f_2}t(t, y) = \DD Xt(t) = 1.
$$
In each of these expressions, we consider only the parameters of the functions when taking derivatives, so there is no ambiguity.
The expression $\DDf{f_1}t$ and $\PDf{f_2}x$ don't even make sense. When we write something like "compute $\DDf ft$ given $f(x, y) = x + y$ and $x(t) = t$", this is an implicit way asking for $\PDf{f_2}t$. It's saying "differentiate with respect to $t$, holding $y$ constant, and treat $x$ as if it depends on $t$".
In an expression like $\nabla\times\mathbf v$, what we want is how $\mathbf v$ varies through space at an instant in time. If $\mathbf v$ is the velocity of a particle, the expression $\nabla\times\mathbf v$ doesn't even make sense because there is no "velocity throughout space" that can vary at an instant in time when there is only one particle. Put another way, if $\mathbf v = v(t)$, then what does $\PDf{\mathbf v}x$ mean? If we did force spacial dependence on $\mathbf v$, then this is how I would formalize it:
let $(X(t), Y(t), Z(t))$ be the coordinates of a particle which vary with time. To be very clear, I will use $X', Y', Z'$ for their derivatives. Then we define
$$
  \mathbf v(x, y, z, t) = \begin{cases}
    (X'(t), Y'(t), Z'(t)) &\text{if }(x, y, z) = (X(t), Y(t), Z(t)) \\
    (0, 0, 0) &\text{otherwise}.
  \end{cases}
$$
But if the velocity isn't $(0,0,0)$ for all $t$, this is not a differentiable function, nor is it partial differentiable except for $\PDf{\mathbf v}t$. Hence $\nabla\times\mathbf v$ doesn't even exist.
So when $\nabla\times\mathbf v$ is meaningful has to be when we have a velocity field, such as with a fluid, or an extended rigid body, or many particles buzzing around all at once.
However, that's all about $\nabla\times\mathbf v$ being a standalone expression; that isn't quite what's going on in $\nabla(\dot{\mathbf x}\cdot\mathbf A)$. This expression is meant to be interpreted as a function
$$
  f(\mathbf x, t) = \Bigl[\nabla_{\mathbf w}(\mathbf X'(t)\cdot\mathbf A(\mathbf w, t))\Bigr]_{\mathbf w = \mathbf x}
$$$$
  \mathbf w = (w_1, w_2, w_3),\quad \nabla_{\mathbf w} = \left(\PD{}{w_1}, \PD{}{w_2}, \PD{}{w_3}\right),
$$
where $\mathbf X$ is the function describing the position of the particle. Note that $\mathbf X$ has nothing to do with $\mathbf x$! In this recasting, I've made it explicit what is varying: $\mathbf w$ is what's varying, all else is held constant, and once the differentiation is said and done we substitute $\mathbf x$ for $\mathbf w$. When we go to simplify $f$, then with $t$ held constant we get two terms
$$
  (\mathbf A(\mathbf w, t)\cdot\nabla_{\mathbf w})\mathbf X'(t),\quad
  \mathbf A(\mathbf w, t)\times(\nabla_{\mathbf w}\times\mathbf X'(t)),
$$
but we're differentiating a constant in each case, which yields $0$. So what the derivation you gave shows is that the function $f$ can be written as
$$
  f(\mathbf x, t) = \Bigl[(\mathbf X'(t)\cdot\nabla_{\mathbf w})\mathbf A(\mathbf w, t)\Bigr]_{\mathbf w=\mathbf x} + \mathbf X'(t)\times\Bigl[\nabla_{\mathbf w}\times\mathbf A(\mathbf w, t)\Bigr]_{\mathbf w=\mathbf x}.
$$

Edit:
Upon rereading this, there is one thing I want to add. When we want to interpret the function $f(\mathbf x, t)$ above, what we want is to evaluate $f(\mathbf X(t), t)$ since we want to identify the parameter $\mathbf x$ with the position of the particle $\mathbf X(t)$. You might be tempted to define a functional
$$
  F : (\R^3)^\R \times \R \to \R,
$$$$
  F(\mathbf X, t) = \Bigl[\nabla_{\mathbf X(s)}(\mathbf X'(s)\cdot\mathbf A(\mathbf X(s), t))\Bigr]_{s=t},
$$
where $(\R^3)^\R$ is the space of functions $\R \to \R^3$, and the intent is that after applying $\nabla_{\mathbf X}$ the resulting fuction is evaluated at $t$. This would let us pass around the particle position function $\mathbf X(t)$ directly. But now we have to define what a gradient with respect to a function is, and how to differentiate the derivative operator $\mathbf X \mapsto \mathbf X'$, and this a much more fraught endeavor than just using the $f(\mathbf x, t)$ interpretation.
A: You are confusing the velocity of the particle $\dot{\boldsymbol x}$ with the fluid velocity field $\boldsymbol u$. This comes from the double usage of the symbol $\boldsymbol x$. One describes the path of a particular particle, the other is a placeholder for an arbitrary position in space.
The particle velocity $\dot{\boldsymbol x}(t)$ is the velocity of a given particle at a time $t$. The velocity field $\boldsymbol u$ is an abstract vector field that describes the velocity of a fluid at any given point in space. This is already enough to answer your question.

Examining this idea will already produce the general form of the Navier-Stokes equations. Consider a velocity field $\boldsymbol u$, depending both on time and space, and a moving particle, having position $\boldsymbol x(t)$ at time $t$.
The force felt on the particle is some external force $\boldsymbol F$, that, in general, can depend on time and space, plus the divergence of the Cauchy stress tensor $\boldsymbol \sigma$, which is a tensor field that depends in general on time and space as well. That is to say,
$$\ddot{\boldsymbol x}(t)=(\nabla\cdot\boldsymbol \sigma)(t,\boldsymbol x(t))+\boldsymbol F(t,\boldsymbol x(t))\tag{1}$$
However, we are aware that the velocity of the particle must be the same as the value of the velocity vector field $\boldsymbol u$ at the given position and time:
$$\dot{\boldsymbol x}(t)=\boldsymbol u(t,\boldsymbol x(t))$$
Now take the time derivative of both sides:
$$\frac{\mathrm d}{\mathrm dt}\dot{\boldsymbol x}(t)=\ddot{\boldsymbol x}(t)=\frac{\mathrm d}{\mathrm dt}\big(\boldsymbol u(t,\boldsymbol x(t))\big)$$
Using some chain rule, this expands as follows:
$$\frac{\mathrm d}{\mathrm dt}\big(\boldsymbol u(t,\boldsymbol x(t))\big)=\frac{\partial \boldsymbol u}{\partial t}(t,\boldsymbol x(t))+\sum_{i}\frac{\mathrm d x_i(t)}{\mathrm dt}~\frac{\partial \boldsymbol u}{\partial x_i}(t,\boldsymbol x(t)) $$
Recognizing this as a dot product,
$$ \ddot{ \boldsymbol x}(t)=\frac{\partial \boldsymbol u}{\partial t}(t,\boldsymbol x(t))+\dot{\boldsymbol x}(t)\cdot \nabla\boldsymbol  u(t,\boldsymbol x(t))$$
Replacing $\dot{\boldsymbol x}$ with $\boldsymbol u$,
$$\ddot{ \boldsymbol x}(t)\\=\frac{\partial \boldsymbol u}{\partial t}(t,\boldsymbol x(t))+\boldsymbol u(t,\boldsymbol x(t))\cdot \nabla\boldsymbol  u(t,\boldsymbol x(t))\\ =(\nabla\cdot\boldsymbol \sigma)(t,\boldsymbol x(t))+\boldsymbol F(t,\boldsymbol x(t))$$
Because we want this to be true for any trajectory of any particle through the fluid, we replace the specific trajectory $\boldsymbol x(t)$ with a generic position $\boldsymbol x$ and proceed to drop the arguments on both sides, leading to the general form of the momentum equation:
$$\frac{\mathrm D\boldsymbol u}{\mathrm Dt}:=\frac{\partial \boldsymbol u}{\partial t}+\boldsymbol u\cdot \nabla \boldsymbol u=\nabla\cdot\boldsymbol \sigma+\boldsymbol F\tag{2}$$
While $(1)$ is a equation about a particular particle, equation $(2)$ is pertaining to an entire velocity field.
