Step 1: Spot what is being oxidised and what is being reduced.
Let us look at the players. Manganese starts in $MnO_4^-$ and ends as $Mn^{2+}$, so its oxidation number falls. Iodine starts as $I^-$ and ends in $IO_3^-$, so its oxidation number rises. The one that falls is reduced (the oxidant), the one that rises is oxidised (the reductant).
Step 2: Count the change for manganese.
In $MnO_4^-$, oxygen is $-2$, so $x + 4(-2) = -1$ gives $x = +7$. In $Mn^{2+}$ it is $+2$. So manganese drops from $+7$ to $+2$, a gain of $5$ electrons per atom.
Step 3: Count the change for iodine.
In $I^-$ iodine is $-1$. In $IO_3^-$, $x + 3(-2) = -1$ gives $x = +5$. So iodine climbs from $-1$ to $+5$, a loss of $6$ electrons per atom.
Step 4: Use the simple electron-balance shortcut.
Instead of writing full half-reactions, just make total electrons lost equal total electrons gained. One $Mn$ gains $5$, one $I$ loses $6$. To match, multiply $Mn$ by the iodine's number and $I$ by the manganese's number. The least common multiple of $5$ and $6$ is $30$.
Step 5: Assign the coefficients.
We need $30$ electrons exchanged, so we need $\dfrac{30}{5} = 6$ manganese units and $\dfrac{30}{6} = 5$ iodide units. That gives $x = 6$ for $MnO_4^-$ and $y = 5$ for $I^-$.
Step 6: Sanity check with the full equation.
Plugging these in, the balanced reaction reads \[ 6MnO_4^- + 5I^- + 18H^+ \rightarrow 6Mn^{2+} + 5IO_3^- + 9H_2O \] Charge on the left is $6(-1)+5(-1)+18(+1) = +7$ and on the right $6(+2)+5(-1) = +7$, and atoms balance too. So the ratio we want is $x:y = 6:5$.
\[ \boxed{6:5} \]