Does A5 have a subgroup of order 6?

If $\pi\in S_5$ is a permutaion of $\{1,2,3\}$, then either $\pi$ or $\pi\circ (4\,5)$ is an even permutation, i.e., $\in A_5$. This allows us to embed $S_3\to A_5$.

Or have fun with geometry: $A_5$ is the symmetry group of the dodekahedron. It is possible to select $4$ of its $20$ vertices that make up a regular tetrahedron $T_1$. A rotation of the dodekahedron by $72^\circ$ turns $T_1$ into another such tetrahedron $T_2$. Then the subgroup of $A_5$ that fixes $T_1\cup T_2$ turns out to be of isomorphic to $S_3$ (why?)

Although you said that you didn't want a yes/no answer, my hint to you is that the answer is yes, so you should just try to construct a subgroup, $H$, of order 6. Look for elements $\sigma$ and $\rho$ of order 2 or 3. A priori there are two possibilities for $H$ - namely $\mathbb Z_6=\langle\rho, \sigma | \sigma \rho \sigma ^{-1}=\rho, \quad \rho^3=e=\sigma^2\rangle$ and $D_6=\langle\rho, \sigma | \sigma \rho \sigma ^{-1}=\rho^{-1}, \quad \rho^3=e=\sigma^2\rangle. $ $\mathbb Z_6$ has an element of order 6, but no such element exists in $A_5$ - you'd need a product of 2-cycles and a 3-cycles, but since you're only allowed to permute 5 objects, you can only get elements like $(12)(345)$ which are odd, so not in $A_5$. So the only possibility is $D_6$, but if you choose your elements $\sigma$ and $\rho$ well you'll get this.