# Does time speed up or slow down near a black hole?

To expand on Javier's answer, the symbol $$\tau$$ represents proper time, i.e. time as measured in a particular reference frame. You're using this symbol for both the proper time in the frame of the stationary observer and the proper time in the frame of the orbiting observer. Equating these two different proper times is incorrect, which is why you are getting the wrong answer. The quantity that does not change between frames is $$t$$, which is the coordinate time. In this case, we've defined the coordinates so that $$t$$ is the time as measured by the observer at infinity.

To avoid confusion, let's use $$\tau_\infty$$ for proper time in the frame of the stationary observer and $$\tau_{orbit}$$ for proper time in the frame of the orbiting observer. Then as you correctly showed,

\begin{align} d\tau_\infty^2 &= dt^2 \\ d\tau_{orbit}^2 &= \left( 1-\frac{3GM}{r_0} \right) dt^2 \end{align}

It follows that

$$d\tau_{orbit}^2 = \left( 1-\frac{3GM}{r_0} \right) d\tau_\infty^2$$

which is the correct answer (as a sanity check, the time as measured by the orbiting observer is less than the time measured at infinity).

You are mixing the two times up. Proper time $$\tau$$ is always the time as measured by the observer you're considering, in this case the orbiting observer, and $$t$$ is coordinate time, which for the Schwarzschild metric is proper time for an observer at infinity. So in a given interval of coordinate time, the orbiting observer measures less time, which means that their clock runs slower.