Consider a cubic, f(x)=a x^3+b x^2+c x+4, a, b, c ∈ R and f^...

$$\begin{aligned} & f^{\prime}(x)=3 a x^2+2 b x+c \quad \Rightarrow \quad f^{\prime}(0)=c=3 \\ & f^{\prime \prime}(x)=6 a x+2 b \Rightarrow f^{\prime \prime}\left(\frac{-2}{3}\right)=-4 a+2 b=0 \Rightarrow b=2 a\end{aligned}$$f^{\prime}\left(\frac{-2}{3}\right)=3 a \times \frac{4}{9}+4 a \cdot\left(\frac{-2}{3}\right)+3=\frac{5}{3}$$\begin{aligned} & \Rightarrow \frac{4 a}{3}=\frac{4}{3} \Rightarrow a=1, b=2 \\ & \therefore \quad f(x)=x^3+2 x^2+3 x+4\end{aligned}$$\frac{d}{d x}(g(x) \cdot f(g(x)))=\frac{d}{d x}(x g(x))=x g^{\prime}(x)+g(x)$$\begin{aligned} & \because \quad f(0)=4 \Rightarrow g(4)=0 \\ & \text { and } g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \Rightarrow g^{\prime}(4)=\frac{1}{f^{\prime}(0)}=\frac{1}{3}\end{aligned}$$\left.\therefore \quad(g(x) \cdot f(g(x)))^{\prime}\right|_{x=4}=\frac{4}{3}$