Show that geometry has local inertial frames

[StevieMurray]

Homework Statement

ds^2 = g_{tt} dt^2 + g_{tx} (dt dx + dx dt)
with g_{tt} = -x and g_{tx} = 3

"Show that this is indeed a spacetime, in the sense that at every point, in any coordinates, the matrix g_{\mu \nu} can be diagonalized with one positive and one negative entry. Hint: You can diagonalize the matrix, but that is the hard way. There is a one-line answer."

Homework Equations

Christoffel Symbols:
g_{\alpha \delta} \Gamma^{\delta}_{\beta \gamma} = \frac{1}{2} \left( \frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} + \frac{\partial g_{\alpha \gamma}}{\partial x ^{\beta}} - \frac{\partial g_{\beta \gamma}}{\partial x^{\alpha}} \right)

ds^2 = g_{\mu \nu}(p) x^{\mu} x^{\nu}

The Attempt at a Solution

I think I have all the components for the solution but I can't put the pieces together. I think I'm supposed to show that the geometry defined in 1. has local inertial frames. Can I do this by showing that the Christoffel symbols vanish (I don't think they all do)? And how do I show that this is valid for any coordinates?
I've also considered expanding the metric as described here: http://mathpages.com/rr/s5-07/5-07.htm :
g_{\mu \nu}(p + x) = g_{\mu \nu}(p) + g_{\mu \nu,\alpha} x^{\alpha} + \frac{1}{2} g_{\mu \nu,\alpha \beta} x^{\alpha} x^{\beta} + \cdots
with
g_{\mu \nu, \alpha} = \left( \frac{\partial g_{\mu \nu}(x)}{\partial x^{\alpha}} \right)_p
and
g_{\mu \nu, \alpha \beta} = \left( \frac{\partial^2 g_{\mu \nu}(x)}{\partial x^{\alpha} \partial x^{\beta}} \right)_p
and then take x to be vanishingly small so that we only use the first term in
ds^2 = g_{\mu \nu}(p) x^{\mu} x^{\nu}
In that case g_{\mu \nu} is just a matrix with numbers which can be diagonalized.

The one-line answer probably has something to do with the fact that some specific derivatives vanish but I don't understand why this is the answer.

 

[Oxvillian]

Maybe det g can tell you something useful?