The function of the oxygen isotope is to determine what mechanism the reaction undergo. The mechanism of ether cleavage is SN1 or SN2 depending on the nature of the carbons bonded to oxygen. If both carbons are primary, cleavage involves a SN2 reaction, the halide ion being the nucleophile. With secondary or tertiary carbons, the $\ce{C-O}$ bond is cleaved by an SN1 mechanism.
In your case you have primary and tertiary carbons bonded to oxygen, but the tertiary carbons are particularly susceptible to cleavage by acid. So your reaction with one equivalent of $\ce{HI}$ will undergo cleavage with SN1 mechanism, forming $\ce{(CH3)3CI}$ and $\ce{H3COH}$.
The SN1 or SN2 mechanism starts with the formation of the oxonium cation:

And then, the characteristics of the carbon atoms bonded to the oxygen select between the SN1 or SN2. If the mechanism is SN2 the nature of the nucleophile also is important, otherwise not. The solvation energy, the strength of the bond with carbon, electronegativity, polarizability and sterics involving the nucleophile are considering factors.
Taking the case if the reaction undergo SN2; the transition state would have a partial $\ce{H3C-I}$ bond and a partial $\ce{H3C-O}$ bond. The free energy of this transition state is higher in energy because the $\ce{C-I}$ bond is weak ($\pu{213 kJ/mol}$), compared to $\ce{C-O}$ ($\pu{336 kJ/mol}$). In general the $\ce{C-I}$ bond is the weaker of the 14-17 groups general elements in organic chemistry (the source is a table from the organic chemistry class that I had). So, the activation energy of the SN2 mechanism is high.
However, if the reaction is SN1, the relative nucleophilicity has no effect on the rate of the reaction. Moreover, the substitution proceeds by rate-determining heterolytic dissociation of the reactant to a carbocation and the leaving group. So, the stability of the carbocation, the nature of the leaving group and the solvent's ionizing power is important. The electronic effect that stabilizes is the electron release, the accepting ability of the leaving group and the capacity of the solvent to stabilize the charge separation in the transition state. The mechanism would be:

where, like I said in the comment, the $\ce{-O(H+)CH3}$ part of the oxonium cation is a good leaving group. For the other hand, the carbocation intermediate is tertiary, i.e. the most stable carbocation. This stability is due to hyperconjugation.