= Solution
For a point charge, the <charge density> and <current density> are
$$
\rho(x',s)=q\delta^{(3)}(x'-y(s)),
\qquad
J(x',s)=qv(s)\delta^{(3)}(x'-y(s)).
$$
Insert these into the <retarded electromagnetic potential>. The argument of the <Dirac delta function> depends on $x'$ through
$$
s=t-\frac{|x-x'|}{c}.
$$
At its unique zero $x'=y(t_{\rm ret})$, the <Jacobian determinant> contributes the factor
$$
\left|1-\frac{n\cdot v}{c}\right|^{-1}.
$$
Since the source is subluminal, this quantity is positive. Writing $\mathbf R=x-y(t_{\rm ret})$, $R=|\mathbf R|$, and evaluating $\mathbf v$ at $t_{\rm ret}$ gives the <Liénard–Wiechert potentials>
$$
\boxed{
\phi(x,t)=\frac{q}{4\pi\epsilon_0}
\frac1{R-\mathbf v\cdot\mathbf R/c},
\qquad
\mathbf A(x,t)=\frac{\mu_0q}{4\pi}
\frac{\mathbf v}{R-\mathbf v\cdot\mathbf R/c}.}
$$
The identity $\mu_0\epsilon_0c^2=1$ equivalently gives $\mathbf A=\mathbf v\phi/c^2$.
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