$\begin{aligned} & \quad f(x)=\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x} \\ & \text { and } \quad g(x)=\frac{1}{1+\frac{1}{f(x)}}=\frac{1}{1+\frac{1+x}{x}}=\frac{x}{2 x+1} \\ & \therefore \quad g^{\prime}(x)=\frac{(2 x+1) \cdot 1-x(2)}{(2 x+1)^2}=\frac{1}{(2 x+1)^2} \\ & \text { Now, } g^{\prime}(2)=\frac{1}{(4+1)^2}=\frac{1}{25}\end{aligned}$