Like you already suspect a DMM won't give you necessarily the answer. If the only levels you read are only 3.3V (capital "V", since it's derived from a person's name) and 0V it's a safe bet that it's digital. A PWM signal is also digital, but will show a voltage between 3.3V and 0V on a DMM, since it includes a low-pass filter. The DMM won't tell an analog and PWM signal apart. You'll need an oscilloscope to actually see what the signal looks like.
If you can trace the signal back to the output of a microcontroller, without filtering, it's most likely a digital signal. While most microcontroller families are available with on-chip ADC, very few also have DACs on board.
I've designed a small system that might work for your particular application. Here is the schematic:
I forgot the reference node, it must be connected to terminal 2 of BT1.
How does this work?
From high to low
First of all let's assume we can neglect the current flowing in/from "low".
When high is pulled up (5V) in R1 no current flows, while the three diodes are conducting. Assuming a forward drop voltage of 0.6V the voltage at low will be 3.2V, and the current flowing from high will be approximately 320uA.
When high is pulled down (0V) all the diodes are interdicted, so the voltage at low will be pulled down by R2. The current that high must sink is approximately 500uA.
From low to high
Now let's assume high is not consuming current.
When low is pulled up (3.3V) the diodes can not conduct because the voltage at high would be more than 5V, so high is pulled up by R1, the diodes are off and low must provide about 330uA.
When low is pulled down (0V) the diodes are correctly polarized, R2 has zero volt across, the voltage at high is about 1.8V and the current sunk by low is approximately 180uA.
As you can see, the big problem is that 1.8V is a bit too much: a CMOS circuit would probably read that "low", while a TTL is likely to read that "high". A better approach could use a 1.5V zener diode instead of the three small signal diodes, with the cathode connected to R1 and the anode to R2. The resistor will probably need to be reduced to meet the minimum polarization current of the zener diode.
One last thing about the resitors is that you can use any value from 1k to 100k, of course higher resistance values correspond to lower current consumption, but also to slower transient response, and vice versa.
Best Answer
The cheapest way is to use a voltage divider consisting of two resistors. The most robust way is to use a level converter IC (SN74LVC1T45 or similar). These have a VCCA and VCCB: Set VCCA to 3.3V, and VCCB to 5.0V, and the IC will translate signals between the two VCC domains.