Given differential equation
$(x-2 y+1) d y-(3 x-6 y+2) d x=0$
$\Rightarrow \quad \frac{d y}{d x}=\frac{3(x-2 y)+2}{(x-2 y)+1}$
Put $x-2 y+1=t \Rightarrow 1-2 \frac{d y}{d x}=\frac{d t}{d x}$
$\therefore \quad \frac{1}{2}\left(1-\frac{d t}{d x}\right)=\frac{3(t-1)+2}{t}=\frac{3 t-1}{t}$
$\Rightarrow \quad \frac{1}{2}-\frac{3 t-1}{t}=\frac{1}{2} \frac{d t}{d x} \Rightarrow \frac{t-6 t+2}{2 t}=\frac{1}{2} \frac{d t}{d x}$
$\Rightarrow \quad \int \frac{t}{2-5 t} d t=\int d x \Rightarrow-\frac{1}{5} \int \frac{5 t-2+2}{5 t-2} d t=\int d x$
$\Rightarrow \quad-\frac{1}{5}\left[t+\frac{2}{5} \ln |5 t-2|\right]=x+C^{\prime}$
$\Rightarrow-\frac{x-2 y+1}{5}-x+\frac{2}{25} \ln |5 x-10 y+3|=C^{\prime}$
$\Rightarrow \quad \ln (5 x-10 y+3)^{2 / 25}=C^{\prime}-\frac{1}{5}-\frac{1}{5}(6 x-2 y)$
$\Rightarrow \quad(5 x-10 y+3)^{2 / 25}=\frac{e^{C^1-\frac{1}{5}}}{e^{1 / 5(6 x-2 y)}}$
$\Rightarrow\left(x-2 y+\frac{3}{5}\right)^{2 / 25} e^{\frac{1}{5}(6 x-2 y)}=C$