For a rocket travelling to a planet, which reference frame measures the dilated time? I'm getting into some confusion with relativity. 
Consider a rocket and the Earth as two observers, and then the rocket travels to a distant star. Is the proper time taken measured in the frame of the rocket or the Earth? 
As far as I know the rocket is at rest relative to the event, and the Earth is moving at a velocity $v$ relative to the event, so maybe I'm just getting confused about the definitions of proper time and dilated time. 
If the Earth measures the dilated time then is the time it measures given by $t=t_0\gamma$, with $t_0$ being the proper time? And the proper time would be the time as measured by the rocket, so simply $t_0=d/v$?
 A: I think you're kind of misunderstanding the notion of what proper time and dilated time is.
Special relativity, at base, is a theory of four dimensional spacetime.  It is always clearest when you think of things in terms of different points in spacetime, and paths between those points.  
So, if you have your spaceship travelling from earth to mars, you can think of three spacetime events -- the rocket leaving earth (point O), the rocket arriving at mars (point M), and the time when the people on Earth observe the rocket as arriving at mars (let them subtract out the light travel time) (point E).  
There are two paths through spacetime here, the path taken by the rocket (OM), and the path taken by the people on Earth (OE).  Each of these paths have a proper time associated with them.  They are both "proper" times.  The point, though, is that they won't be the SAME time interval -- they will be dilated RELATIVE to each other.
A: "If the Earth measures the dilated time then is the time it measures given by $t=t_0 \gamma$, with $t_0$ being the proper time? And the proper time would be the time as measured by the rocket, so simply $t_0 = d/v$?"
If $d$ is meant to be the distance from Earth to the star in their own mutual rest frame, then $d/v$ would be the coordinate time in that same frame for a rocket moving at $v$ to move from Earth to the star, i.e. $t=d/v$ where $t$ is the time in the Earth frame. If $t_0$ is the proper time along the rocket's worldine from the event of leaving Earth to arriving at the star, then the correct distance/velocity relation would be $t_0 = d' /v$, where $d'$ is the distance between the Earth and the Star in the rocket's rest frame, which is shorter than the distance between them in the Earth/Star rest frame, due to length contraction. In general, whenever you want to use an expression of the form time = distance/velocity, you have to make sure that all three quantities are defined in the same frame (and keep in mind that proper time between events on an inertial observer's worldline is identical to the coordinate time between those events in the observer's rest frame, so $t_0$ is the coordinate time in the rocket's rest frame). Since you said this is homework, I'll let you play around with the equations to see how it can be true that $t=d/v$, $t_0 = d'/v$, and that $t=t_0 \gamma$.
