Find short location on PCB
A surer method is to apply current and then measure voltage drop across sections of track with a sensitive voltmeter. Many multimeter (even cheap ones) have a 200 mV (milliVolt) range which, with a 4 digit display gives 100 uV (microVolt) resolution (not accuracy, but that's another issue). If you pass say 100 mA along a track then you need 1 milliOhm of track to drop 100 uV. 


\$(R = \frac{V}{I} = \frac{0.0001}{0.1} = 0.001)\$ 


If you place two probes at either end of a track section the polarity of the voltage drop will show you the direction of current flow. You can literally follow the current around the board. You can find the point at which current leaves a track - voltage drop per track length will be smaller or of opposite polarity past a point of current departure. 

If that's not sensitive enough, many meters have a 200 uA range with a resolution of 0.1 uA (!). Due to issues of meter internal resistance and specific implementation one or other of these ranges may work best for you, but either way you have a useful short-circuit-current tracer.  

Note that the voltage range is less liable to be damaged when doing this. When using a current range, if you probe two locations which are not on the same track and which have a significant voltage difference it is easy for far more than 200 uA to flow and, depending on meter circuitry, damage may occur.

You can set the test current to a useful minimum value by using a current limited supply to apply a current to a section of track and then observing the results with your meter. Current may be limited to a value that works well enough with you test equipment.

Whether a given test current is "safe" depends on your circuitry and is not able to be stated for a general situation. If you use a voltage and current limited supply and set the supply voltage to just high enough to produce a short circuit current suitable for testing then you will know the maximum energy that is able  to be dissipated anywhere. For example, if you set Imax to 100 mA and Vsupply to 2 Volts then total available energy = V x I = 2 x 0.1 = 200 mW (milliWatt). That level of dissipation MAY cause problems in specific devices but is not liable to cause thermal damage to almost anything.