$a)$

$\{\begin{array}{l}2x\ne 0\\ 3-x\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne 0\\ x\ne 3\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{0,3\right\}$

$\frac{x-3}{\sout{2x}}\cdot \frac{{\sout{6x^2}}^{3x}}{3-x}=\frac{3x\left(x-3\right)}{3-x}=\frac{3x\left(x-3\right)}{\left(-1\right)\cdot \left(x-3\right)}=\frac{3x}{-1}=-3x$

$x=-1\in D\Rightarrow -3x=-3\cdot \left(-1\right)=3$