Probability that $A \cup B$ = S and $A \cap B = \phi $ Let $S$ be a set containing $n$ elements and we select two subsets: $A$ and $B$ at random then the probability that $A \cup B$ = S and $A \cap B = \varnothing $ is?
My attempt
Total number of cases= $3^n$ as each element in set $S$ has three option: Go to $A$ or $B$ or to neither of $A$ or $B$
For favourable cases: Each element has two options: Either go to $A$ or to $B$ which gives $2^n$ favourable cases.
Is my approach correct?

Assuming you mean that $A$ and $B$ are picked independently and uniformly at random.
Edit: below, I'm answering two different questions. I guess, after reading yours carefully, that what you intended is the second one.
First interpretation of the question: "find $\Pr[A\cap B=\emptyset]$ and  $\Pr[A\cup B=S]$"
Consider a fixed element $s\in S$. The probability that it belongs to neither $A$ nor $B$ is
$$
\Pr[s\notin A \text{ and } s\notin B] = \Pr[s\notin A]\cdot \Pr[s\notin B] = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}
$$
where we used independence for the first equality. We have $A\cup B = S$ if, and only if, all elements belong to at least one of $A$ and $B$, which by the above happens for each fixed element $s$ with probability $1-\frac{1}{4}$. Again by independence, this results in
$$
\Pr[ A\cup B= S] = \Pr[\forall s,\ s\in A\cup B]
= \prod_{s\in S} \Pr[s\in A\cup B] = \left(\frac{3}{4}\right)^n.
$$
Similarly, the probability that a given fixed element $s\in S$ belongs to $A\cap B$ can be shown to be $\frac{1}{4}$. The probability that no element ends in $A\cap B$ is then
$$
\Pr[ A\cap B=\emptyset] = \prod_{s\in S}\Pr[ s\notin A\cap B]
= \prod_{s\in S}\left(1-\Pr[ s\in A\cap B]\right) = \left(1-\frac{1}{4}\right)^n=\left(\frac{3}{4}\right)^n.
$$
Second interpretation: "find $\Pr[A\cap B=\emptyset \text{ and } A\cup B=S]$"
This is similar to the above approach. Fix any element $s\in S$: the probability that $s$ ends up in exactly one of $A$ and $B$ is 
$$\begin{align}
\Pr[ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ]
&=\Pr[s\in A \text{ and } s\notin B] + \Pr[s\notin A \text{ and } s\in B]\\
&=\Pr[s\in A]\cdot\Pr[s\notin B] + \Pr[s\notin A]\cdot\Pr[s\in B]\\
&=\frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2}
=\frac{1}{2}
\end{align}$$
so the probability you want is
$$\begin{align}
\Pr[ \forall s\in S,\ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ]
&=\prod_{s\in S }\Pr[ ( s\in A \text{ and } s\notin B ) \text{ or } ( s\notin A \text{ and } s\in B ) ]\\
&=\left(\frac{1}{2}\right)^n
\end{align}$$
