For an observer who is not in a gravitational field with any relevant strength, the speed of light in a (far enough away) point under the influence of stronger gravitation is not the universial constant $c_0$ but some lower value. For the Schwarzschild Metric I got a nice answer on Physics StackExchange for the observed speed of light.
Basically the same derivation can be done for a general metric $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$ where $dx^0 = c_0 dt$? Divide by $dt^2$ to get $$ -g_{00} c_0^2 = \frac{g_{ij}dx^i dx^j}{dt^2} + 2 c_0 \sum_{i=1}^3 \frac{g_{0i} dx^i}{dt} \qquad i,j\in\{1,2,3\} $$ using the fact that the metric is symmetric, hence the simple sum times $2$.
Now comes the same trick as in the StackExchange answer cited above. The observed velocity $c$ in an arbitrary direction is $c \vec{e}$ where $|\vec{e}| = |(e_1, e_2, e_3)| = 1$ is a unit vector and $dx^i/dt = c e_i$. So we somewhat trivially get: $$ -g_{00} c_0^2= g_{ij}e^i e^j c^2 + 2 c_0 c \sum_{i=1}^3 g_{0i} e^i \qquad i,j\in\{1,2,3\} $$ With $a = g_{ij}e^i e^j$ and $b = \sum_{i=1}^3 g_{0i} e^i$ this is \begin{align*} -g_{00} c_0^2 &= a c^2 + 2 c_0 c b \\ -g_{00} c_0^2 /a &= c^2 + 2 c_0 c b/a \\ -g_{00} c_0^2 /a + (c_0 b/a)^2 &= (c_0 b/a + c)^2 \\ c &= \sqrt{ -g_{00} c_0^2 /a + (c_0 b/a)^2 } - c_0 b/a \\ &= c_0 \sqrt{ -g_{00} /a + (b/a)^2 } - c_0 b/a \\ %&= c_0/a \left( \sqrt{ -g_{00} a + b^2 } - b\right) \end{align*} Dragging along the $c_0$ and the $c$ instead of just working with $c_0=1$ allows at least a simple unit check, and yes, both sides are scalar velocities.
Another quick test: In the Minkowski Metric, the matrix with the diagonal $(1, -1, -1, -1)$, we get $a =g_{ij}e^ie^j = -1 \sum_{j=1}^{3} (e^j)^2 = -1$ and $b=0$, so we have $c = c_0 \sqrt{-1 / -1} =c_0$, as required.