The problem is not possible.

That means 0 = S(n-2) - 3/S(n-1)

so S(n-1)*S(n-2)=3

But S(n-1) = S(n-3) - 3/S(n-2)

and substituting in gives

[S(n-3) - 3/S(n-2)]*S(n-2)=3

S(n-3)*S(n-2)=0  <--- Algebra error.  This should be 6.

Now S(n-2) cannot be 0, so S(n-3)=0

This means S(n-1)=0-3/S(n-2)

meaning S(n-2)*S(n-1)=-3

Which is a contradiction.

So there is no n such that S(n)=0.

***** Edit *****

Daniel's solution seems valid and a quick check with Excel seems to confirm this.  We posted at nearly the same time so I didn't see his until I hit enter.  I think something must be wrong with the above.  I wonder what.

Now that I've got S(n-3)*S(n-2)=6 I could probably show S(n-(a+1))*S(n-a)=3a and work backwards to give the same solution as Daniel.