= Solution
Put
$$
B^2=\frac{(1-A)^2}{(1+A)^2},
\qquad C=(1+A)^4.
$$
The rest-frame metric can be written as $C\,d\bar s_{\rm Minkowski}^2+(C-B^2)d\bar t^2$. The Minkowski term is invariant under the <Lorentz transformation>, while $d\bar t=\gamma(dt-vdx)$. Therefore
$$
\boxed{f(A)=(1+A)^4,
\qquad
g(A)=(1+A)^4-\frac{(1-A)^2}{(1+A)^2},}
$$
and
$$
\boxed{A=\frac{m}{2\sqrt{\gamma^2(x-vt)^2+y^2+z^2}}.}
$$
These substitutions give precisely
$$
ds^2=f(A)(-dt^2+dx^2+dy^2+dz^2)+\gamma^2g(A)(dt-vdx)^2.
$$
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