$xy-3x+5y=0$
$⇒xy-3x+5y-15=-15$
$⇒y(x+5)-3(x+5)=-15$
$⇒(x+5)(y-3)=-15$
TH1,2:
$\left[ \begin{array}{l}\left \{ {{x+5=-5} \atop {y-3=3}} \right.\\\left \{ {{x+5=5} \atop {y-3=-3}} \right.\end{array} \right.$
$⇒\left[ \begin{array}{l}\left \{ {{x=-10} \atop {y=6}} \right.\\\left \{ {{x=0} \atop {y=0}} \right.\end{array} \right.$
TH3,4:
$\left[ \begin{array}{l}\left \{ {{x+5=-1} \atop {y-3=15}} \right.\\\left \{ {{x+5=1} \atop {y-3=15}} \right.\end{array} \right.$
$⇒\left[ \begin{array}{l}\left \{ {{x=-6} \atop {y=18}} \right.\\\left \{ {{x=-4} \atop {y=18}} \right.\end{array} \right.$
TH5,6:
$\left[ \begin{array}{l}\left \{ {{x+5=-3} \atop {y-3=5}} \right.\\\left \{ {{x+5=3} \atop {y-3=-5}} \right.\end{array} \right.$
$⇒\left[ \begin{array}{l}\left \{ {{x=-8} \atop {y=8}} \right.\\\left \{ {{x=-2} \atop {y-2}} \right.\end{array} \right.$
TH6,7:
$\left[ \begin{array}{l}\left \{ {{x+5=-15} \atop {y-3=1}} \right.\\\left \{ {{x+5=15} \atop {y-3=-1}} \right.\end{array} \right.$
$⇒\left[ \begin{array}{l}\left \{ {{x=-20} \atop {y=4}} \right.\\\left \{ {{x=10} \atop {y=-2}} \right.\end{array} \right.$
Vậy $(x;y)=..........$
