smc

We use mass points for this problem. Let $\text{m}A$ denote the mass of point $A$. Rewrite the expression we are finding as $$\frac{EF}{FC}+\frac{AF}{FD}=\frac{FE}{FC}+\frac{FA}{FD}=\frac{\text{m}C}{\text{m}E}+\frac{\text{m}D}{\text{m}A}$$ Now, let $\text{m}C=2$. We then have $2\cdot 1=\text{m}B\cdot 2$, so $\text{m}B=1$, and $\text{m}D=2+1=3$ We can let $\text{m}A=3$. We have $\text{m}E=\text{m}A+\text{m}B=3+1=4$ From here, substitute the respective values to get $$\frac{\text{m}C}{\text{m}E}+\frac{\text{m}D}{\text{m}A}=\frac{2}{4}+\frac{3}{3}=\frac{1}{2}+1=\frac{3}{2}$$ This answer corresponds to $\boxed{\frac{3}{2}}$.