Given, slope of tangent line to curve at (x,y) is ‌
xy2+y
x
i.e., ‌‌‌
dy
dx
=‌
xy2+y
x
⇒‌‌
dy
dx
=y2+‌
y
x
⇒xdy=xy2dx+ydx ⇒‌xdy−ydx=xy2dx⇒‌‌
xdy−ydx
y2
=xdx Integrating both sides, we get ‌
−x
y
=‌
x2
2
+C . . . (i) The curve intersect line at x=−2 Then, x=−2, is satisfied by x+2y=4 Hence, (−2)+2y=4 Gives, y=3 ∴ Curve passes through (2,−3). Use (2,−3) is Eq. (i), we get ‌
−2
−3
=‌
(−2)2
2
+C⇒C=‌
−4
3
∴ The curve is ‌
−x
y
=‌
x2
2
=‌
−4
3
. . . (ii) It also passes through (3,y). Put (3,4) in Eq. (ii), we get ⇒‌‌‌