If we are counting ordered pairs $(X, Y)$, then for each element of the set we have three choices. Put it in set X, in Y or in none of them. So total ways = $3^n$.
If we are counting unordered pairs $(X, Y)$, then except for the pair $({}, {})$, all pairs have been counted twice. So toal ways are $\frac{3^n - 1}{2} + 1$.
Here $n = 6$, so answer for first case is $3^6 = 729$ and for second case $\frac{3^6 - 1}{2} + 1 = 365$.
Another method:
Suppose $X$ has 0 elements (which can be chosen in $\binom{n}{0}$ ways), then $Y$ can include or not include any of the $n$ elements of the give set.
Number of ways = $\binom{n}{0}2^n$
If $X$ has 1 element (which can be chosen in $\binom{n}{1}$ ways), then $Y$ can include or not include any of the remaining $n-1$ elements.
Number of ways = $\binom{n}{1}2^{n-1}$
and so on...
So final answer is $\sum_{i=0}^n \binom{n}{i}2^{n-i} = 3^n$
