Question

=y(y^2+1) dx + x (y^2-1) dy=0

Answer 1

the extra = sign is confusing, but suppose we make it

-y(y^2+1) dx + x (y^2-1) dy=0
then
x(y^2-1) dy = y(y^2+1) dx
(y^2-1)/(y(y^2+1)) = 1/x dx
(2y/(y+2+1) - 1/y) dy = 1/x dx
ln(y^2+1) - lny = lnx + lnc
(y^2+1)/y = cx

If I got that wrong, make the fix and solve in a similar manner