jmc

We have that $$\frac{1}{S_n}=\frac{S_{n-2}+S_{n-1}}{S_{n-2}\cdot S_{n-1}}=\frac{1}{S_{n-1}}+\frac{1}{S_{n-2}}.$$Accordingly, let $T_n=\frac{1}{S_n}.$ Then $T_1=1,$ $T_2=1,$ and $$T_n=T_{n-1}+T_{n-2}$$for $n\ge 3.$ Then $T_3=2,$ $T_4=3,$ $\dots ,$ $T_{12}=144,$ so $S_{12}=\boxed{\frac{1}{144}}.$