This is one answer (seven others are possible by rotation and reflection)   

  

Put the 1 in the middle so that reciprocals can go opposite each other.

Try placing the 6 at a corner:
 $\frac 16$ must go opposite $6$
$\frac12$ and $\frac13$ must go next to $6$ to multiply to $1$ 


But $\frac13$ is opposite $\frac16$ and there is nothing large enough to multiply by to get $1$

$\frac12\times\frac16$ is also too small so swapping $\frac12$ and $\frac13$ would not help



Try placing the $6$ in the middle of an edge: