Given, $y=x \log \left(\frac{x}{2-3 x}\right)$, for $0 < x < \frac{2}{3}$
On differentiating w.r.t. to ' $x$ ', we are getting
$$
\begin{aligned}
\frac{d y}{d x} & =x\left(\frac{1}{x}-\frac{-3}{2-3 x}\right)+\log \left(\frac{x}{2-3 x}\right) \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{2}{2-3 x}+\log \left(\frac{x}{2-3 x}\right)
\end{aligned}
$$
Again differentiating w.r.t. to ' $\chi$ ', we are getting
$$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{-2}{(2-3 x)^2}(-3)+\left(\frac{1}{x}-\frac{-3}{2-3 x}\right) \\
& =\frac{6}{(2-3 x)^2}+\frac{2}{x(2-3 x)}
\end{aligned}
$$
So,
$$
\begin{aligned}
& \frac{d^2 y}{d x^2}\left(\text { at } x=\frac{1}{2}\right)=\frac{6}{(1 / 2)^2}+\frac{2}{1 / 2 \times \frac{1}{2}} \\
& =24+8=32 \\
&
\end{aligned}
$$