Ta xét hàm số $y=\frac{x{{\left( x+1 \right)}^{3}}}{{{\left( {{x}^{2}}+x-6 \right)}^{2}}}$ trên $\left( \frac{1}{2};\frac{3}{2} \right)$ ta có:

$y=\frac{x{{\left( x+1 \right)}^{3}}}{\left( x+3 \right){{\left( {{x}^{2}}+x-6 \right)}^{2}}}=\frac{x{{\left( x+1 \right)}^{3}}}{{{\left( x-2 \right)}^{2}}{{\left( x+3 \right)}^{3}}}>0$$\forall x\in \left( \frac{1}{2};\frac{3}{2} \right)$

$\Rightarrow \ln y=\ln x+3\ln \left( x+1 \right)-2\ln \left( 2-x \right)-3\ln \left( x+3 \right)$

$\begin{align}  & \Rightarrow \frac{y'}{y}=\frac{1}{x}+\frac{3}{x+1}-\frac{2}{x-2}-\frac{3}{x+3}\left( 1 \right) \\  & \Rightarrow \frac{y''.y-{{\left( y' \right)}^{2}}}{{{y}^{2}}} \\ \end{align}$

$=\frac{-1}{{{x}^{2}}}+\frac{-3}{{{\left( x+1 \right)}^{2}}}-\frac{-2}{{{\left( 2-x \right)}^{2}}}-\frac{-3}{{{\left( x+3 \right)}^{2}}}\left( 2 \right)$

Ta có:

$y\left( 1 \right)=\frac{1}{2}\xrightarrow{\left( 1 \right)}y'\left( 1 \right)=\frac{15}{8}\xrightarrow{\left( 2 \right)}y''\left( 1 \right)=\frac{29}{4}$

Vậy đáp án đúng là C.