Measuring li-ion that arduino is running from
Using the 1.1V internal analog reference to measure a draining VCC
source by using a voltage divider on it

You could indeed use a voltage divider, and measure a scaled-down Vcc
against the internal 1.1 V reference. This is, however, not what
the code you posted is doing. It is instead measuring the internal
reference against Vcc, as stated in the comment within the code. The
REFS0 bit selects Vcc as the reference for the ADC. The bits MUX1...MUX3
select the internal reference as the input channel.

result = ADCL;
result |= ADCH&lt;&lt;8;


This idiom dates back from a time when the compiler did not know how to
read 16-bit I/O registers: you had to explicitly read one byte at a time
and then put these together into a 16-bit word. This time is long gone,
and you can now just access the 16-bit register as ADC directly. I
would even get rid of the result variable and just:
return 1126400L / ADC;  // Vcc in mV

Edit: As suggested by the busybee, here is the explanation of the
factor 1126400: The ADC meaures the ratio of its input voltage to its
reference voltage, and returns this ratio scaled by a factor 1024. In
this case,
ADC = 1024 × (Vbg ÷ Vcc)
where Vbg is the voltage of the 1.1 V bandgap reference. Thus,
Vcc = 1024 × Vbg ÷ ADC
If we want Vcc in millivolts, we plus the Vbg voltage in millivolts in
the above equation and get
Vcc (in mV) = 1024 × 1100 ÷ ADC = 1126400 ÷ ADC