You intuition about choosing the bookmaker with the best odds for each outcome is certainly correct. How could it make sense to do otherwise?
So you bet $b_1$ with $B_1$ that $Y$ wins, $b_2$ with $B_2$ that $X$ wins, and $B_3$ with $B_3$ on a draw. Presumably, you have a predetermined amount to bet, so we assume $b_1+b_2+b_3=1$ and interpret the $b_i$ as the fraction bet with each book maker.
Now you want to analyze the profit or loss in the case of each of the three outcomes, and choose the $B_i$ to maximize the profit in the worst case. This is predicated on my understanding of the goal is to make money no matter what the outcome.
EDIT
Out of curiosity, I worked out the details, so I might as well post them. We have the following odds $$\begin{matrix}
&\text{X}&\text{Y}&\text{Draw}\\
\text{B}_1&4.2&\mathbf{2}&3.4\\
\text{B}_2&\mathbf{4.7}&1.9&3.1\\
\text{B}_3&4.1&1.6&\mathbf{3.7}
\end{matrix}$$
where the best odds on each outcome have been bolded.
If we bet $b_i$ with bookmaker $\text{B}_i,\ i=1,2,3$ where we assume $b_1+b_2+b_3=1$ then in order that we not lose money whatever happens, the following conditions must hold. $$\begin{align}
-b_1 +4.7b_2 -b_3 &\geq0\tag{1}\\
2b_1 -b_2 -b_3 &\geq0\tag{2}\\
-b_1 -b_2 +3.7b_3 &\geq0\tag{3}
\end{align}$$
Substituting $b_3=1-b_1-b_2$ in $(1)$ and simplifying gives $$b_2\geq{1\over5.7}\approx .1754$$Similarly, substituting in $(2)$ gives $$b_1=\frac13\approx .3333$$ and substituting in $(4)$ gives $$b_1+b_2\leq{37\over47}\iff b_3\geq{10\over47}\approx.2127$$
The sum of the lower bounds comes to $\approx.7215$ so the problem is feasible. If we lay out about $72\%$ of our capital as computed above, we will never lose, and so the problem is how to bet the remaining $28\%$ to best advantage.
Since this is arbitrage, we want our profit to be the same no matter what the outcome. In order to have the same profit if X wins or there is a draw, we equate the left-hand sides of $(1)$ and $(3)$ and we find$$5.7b_2=4.7b_3\tag{4}$$ To ensure the same profit if there is a draw of if Y wins, we equate the left-hand sides of $(2)$ and $(3)$ and find $$3b_1=4.7b_3\tag{5}$$ Comparing $(4)$ and $(5)$ we see that we must have $$b_1=1.9b_2\tag{6}$$ $(4)$ and $(6)$ now give$$b_2\left(1.9+1+{57\over47}\right)=1\implies b_2={470\over1933}\approx.243145 $$so we have $$\boxed{b_1\approx.461976\\b_2\approx.243145\\b_3\approx.294878}$$
You can, and should, verify that no matter what the outcome, the profit is about $ 38.59\%$ of the amount wagered.
In retrospect, I could have simply solved for the bets that would make the profit the same for all outcomes, and then tested whether it was positive. You can recast the solution that way, if you like.
