It should be easy enough to derive it: we're dealing with an oblate spheroid. The difficulty comes in the choice of coordinates: this page notes the difference between the geocentric and geodetic coordinates (but for some reason the diagram showing the difference is on this page): essentially we have an ellipsoid parametrised as
$$ N(\phi) (\cos{\phi}\cos{\lambda},\cos{\phi}\sin{\lambda},b^2/a^2 \sin{\lambda}), $$
where
$$ N(\phi) = \frac{a^2}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}}} $$
is related the radius at angle $\phi$: we see that it varies from $a$ at the equator to $a^2/b$ at the poles. Now one can compute the metric in the usual way using inner products of tangent vectors: one tedious calculation later, we find
$$ ds^2 = \frac{a^4b^4}{(a^2\cos^2{\phi}+b^2\sin^2{\phi})^3} \, d\phi^2 + \frac{a^4\cos^2{\phi}}{a^2\cos^2{\phi}+b^2\sin^2{\phi}} \, d\lambda^2. $$
As a check, if we put $b=a$, we find this reduces to $a^2 \, d\phi^2 + a^2\cos^2{\phi} \, d\lambda^2$, which also serves as a valuable reminder that $\phi$ is latitude and $\lambda$ is longitude, not the usual coordinates we use in mathematics.
Since geodesics on an ellipsoid are horrible, you're probably better off looking up standard techniques of approximation: that flattening of 1/300 makes things surprisingly messy!
