The idea is straight forward, we just need to test all the numbers. It is possible to optimize the algorithm a little bit by using memoization with the first 2 numbers in the multiplication chain.
The running time of this algorithm is O(n^3) since we test n numbers against all the others n-1 and n-2 numbers in the array.
Another solution is go through the array only once and keep tracking the 3 greatest positive numbers, the 3 smaller negative numbers and the 3 greatest negative numbers and return the result considering if there is any positive number in the array or not. Using this strategy, the running time of the algorithm is O(n) since we only pass in the array once checking the possible results at the end.