I don't understand how the derivative is calculated in this case:
$$ \frac{\partial}{\partial t} \delta\left(\vec{r}-\vec{r}_{i}(t)\right)=\frac{\partial \delta}{\partial x_{i}} \frac{d x_{i}}{d t}+\frac{\partial \delta}{\partial y_{i}} \frac{d y_{i}}{d t}+\frac{\partial \delta}{\partial z_{i}} \frac{d z_{i}}{d t}=-\frac{\partial \delta}{\partial x} \frac{d x_{i}}{d t}-\frac{\partial \delta}{\partial y} \frac{d y_{i}}{d t}-\frac{\partial \delta}{\partial z} \frac{d z_{i}}{d t} $$
The multivariable chain rule is -General rule: Vector-valued functions with multiple inputs The simplest way for writing the chain rule in the general case is to use the total derivative, which is a linear transformation that captures all directional derivatives in a single formula. Consider differentiable functions $ f: \mathbf{R}^{m} \rightarrow \mathbf{R}^{k} $ and $ g: \mathbf{R}^{n} \rightarrow \mathbf{R} $ and a point $ \mathbf{a} $ in $ \mathbf{R}^{n} $. Let $ D_{\mathbf{a}} g $ denote the total derivative of $ g $ at $ \mathbf{a} $ and $ D_{g(\mathbf{a})} f $ denote the total derivative of $ f $ at $ g(\mathbf{a}) $. These two derivatives are linear transformations $ \mathbf{R}^{n} \rightarrow \mathbf{R}^{m} $ and $ \mathbf{R}^{m} \rightarrow \mathbf{R}^{k} $, respectively, so they can be composed. The chain ru for total derivatives is that their composite is the total derivative of $ f \circ g $ at $ \mathbf{a} $ : $ D_{\mathrm{a}}(f \circ g)=D_{g(\mathbf{a})} f \circ D_{\mathrm{a}} g, $
First I have to denote that $\delta$ is interpreted as a function in this case.
I would identify $f=\delta$ and the inner function $g(t)=\vec{r}-\vec{r}_{i}(t)=\vec{r}(x(t),y(t),z(t))-\vec{r}_{i}(x(t),y(t),z(t)) $
So $\delta:\mathbb{R}^3\to \mathbb{R} $ and $g:\mathbb{R}\to \mathbb{R}^3 $, so $\delta \circ g: \mathbb{R}\to \mathbb{R} $ and $D_{\mathrm{t}}(\delta \circ g)=D_{g(\mathbf{t})} \delta \circ D_{\mathrm{t}} g = ...$
with $$D_{\mathrm{t}} g= \dot{\vec{r}}-\dot{\vec{r_i}}= \frac{d x}{d t}\hat{e_x}+\frac{d y}{d t}\hat{e_y}+\frac{d z}{d t}\hat{e_z}-\frac{d x_{i}}{d t}\hat{e_x}-\frac{d y_{i}}{d t}\hat{e_y}-\frac{d z_{i}}{d t}\hat{e_z}=(\frac{d x}{d t}-\frac{d x_{i}}{d t})\hat{e_x}+\frac{d x}{d t}-(\frac{d x_{i}}{d t})\hat{e_y}+(\frac{d y}{d t}-\frac{d z_{i}}{d t})\hat{e_z}$$
$$ D_{g(t)} \delta= D \delta(g(t))=??? $$
Edit- more context: Moving charges form a current $ I $, which is described by the atomic current density $ \vec{J}(\vec{r}, t)=\sum \limits_{i} Q_{i} \vec{v}_{i} \delta\left(\vec{r}-\vec{r}_{i}(t)\right) $ with $ \vec{v}_{i}=\frac{d \vec{r}_{i}}{d t} $ as the velocity of the $ i $-th particle. For charge density and current density, the continuity equation holds: $ \frac{\partial \rho}{\partial t}+\vec{\nabla} \cdot \vec{J}=0$ with the atomac charge density $\rho=\sum \limits_{i} Q_{i} \delta\left(\vec{r}-\vec{r}_{i}\right)$
You typically see those kind of formulas in physics. For example, you can view this as the density of particles for a single particle at position $r_i$. The continuous analogue would be to start with a density $n_0(r)$ and a flow $\phi(r_0,t)$ which gives the new position $r$ at $t$ when starting at $r_0$. Associated with this flow is the velocity field $v(r,t)$ which gives the current velocity of $r$: $$ \partial_t\phi = v\circ \phi \\ v = \partial_t\phi\circ \phi^{-1} $$ You can then compute how the density is carried by the flow: $$ n(r,t) = n_0(\phi^{-1}(r,t)) $$ (the inversion of $\phi$ is only in the spatial part) Assuming differentiability of $n$ and $\phi$, this becomes: $$ \partial_t n = \partial_rn_0 \cdot \partial_t\phi^{-1} $$ Since the reverse flow is obtained by switching the sign of the velocity: $$ \partial_t\phi^{-1} = -v\circ \phi^{-1} $$ you can rewrite it using the velocity field of the flow: $$ \partial_t n = -\partial_rn\cdot v $$ Notice that formula is linear in $n_0$, so it's tempting to decompose it in Dirac deltas. This is how you can interpret your equation. In other terms, it's just way to remember the convective derivative.
Hope this helps.