We have,
$$
\begin{aligned}
& \frac{x}{\left(1+x^2\right)(3-2 x)}=\frac{B x+C}{1+x^2}+\frac{A}{(3-2 x)} \\
& \Rightarrow \frac{x}{\left(1+x^2\right)(3-2 x)}=\frac{B x(3-2 x)+C(3-2 x)+A\left(1+x^2\right)}{\left(1+x^2\right)(3-2 x)} \\
& \Rightarrow x=B x(3-2 x)+C(3-2 x)+A\left(1+x^2\right) \\
& \Rightarrow x=3 B x-2 B x^2+3 C-2 C x+A+A x^2 \\
& \Rightarrow \quad=x=(A-2 B) x^2+(3 B-2 C) x+A+3 C
\end{aligned}
$$
Equating the coefficient of $x^2, x$ and constant term
$$
A-2 B=0,3 B-2 C=1, A+3 C=0
$$
Solving, we get
$$
\begin{aligned}
& A & =\frac{6}{13}, B=\frac{3}{13}, C=-\frac{2}{13} \\
\therefore & C & =-\frac{2}{13}
\end{aligned}
$$