$(1$ points$)$ In an appropriate form, the general solution of $y^{\text{'}}=\frac{(xy+3x-y-3x)}{(xy-2x+4y-8)}$ is

$(x+4{)}^5{e}^y=C(y-3{)}^5{e}^x$

$(x+4{)}^5{e}^y=C(y+3{)}^5{e}^x$

$(x+3{)}^5{e}^y=C(y+5{)}^5{e}^x$

$(x+4{)}^5{e}^x=C(y+3{)}^5{e}^y->$ the right answer, but how?