Chain rule application I want to find $y'$ where $$ y = \frac{\frac{b}{a}}{1+ce^{-bt}}.$$ But I dont want to use quotient rule for differentiation. I want to use chain rule. My solution is:
Write  $$y=\frac{b}{a}\cdot \frac{1}{1+ce^{-bt}}.$$ Then in $$\frac{1}{1+ce^{-bt}},$$ the inner function is $1+ce^{-bt}$ and the outer function is $$\frac{1}{1+ce^{-bt}}.$$
Hence using the chain rule we have 
$$\left(\frac{1}{1+ce^{-bt}}\right)'= \frac{-1}{(1+ce^{-bt})^2} \cdot -bce^{-bt} = \frac{bce^{-bt}}{(1+ce^{-bt})^2}.$$
Thus $$y'= \frac{\frac{b^2}{a}ce^{-bt}}{(1+ce^{-bt})^2}.$$ Am I correct?

Yes.
The outer function is $s\mapsto \displaystyle\frac1s$ or you can call its variable anything. And '$\cdot -bce^{-bt}$' should be in parenthesis: $\cdot (-bce^{-bt})$, else it seems correct.
