= Solution
Put
$$
\rho=\sqrt{r^2+a^2},
\qquad
\varpi=\rho\sin\theta,
\qquad
z=r\cos\theta.
$$
Since $x^2+y^2=\varpi^2$, the Euclidean spatial line element is
$$
dx^2+dy^2+dz^2=d\varpi^2+\varpi^2d\phi^2+dz^2.
$$
Now
$$
d\varpi=\frac r\rho\sin\theta\,dr
+\rho\cos\theta\,d\theta,
\qquad
dz=\cos\theta\,dr-r\sin\theta\,d\theta.
$$
The cross terms cancel. With
$$
\Sigma=r^2+a^2\cos^2\theta,
\qquad
\Delta=r^2+a^2,
$$
we obtain the <oblate spheroidal coordinates> line element
$$
dx^2+dy^2+dz^2
=\frac\Sigma\Delta dr^2+\Sigma d\theta^2
+\Delta\sin^2\theta\,d\phi^2.
$$
Adding the time coordinate gives
$$
\boxed{
ds^2=-dt^2
+\frac{r^2+a^2\cos^2\theta}{r^2+a^2}dr^2
+(r^2+a^2\cos^2\theta)d\theta^2
+(r^2+a^2)\sin^2\theta\,d\phi^2}.
$$
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