Solve dy/dx for sin(xy)=1+x^3-y^2

Answer 1

sin(xy)=1+x^3-y^2

cos(xy) (y + xy') = 3x^2 - 2yy'
y cos(xy) + xcos(xy) y' = 3x^2 - 2yy'
(x cos(xy) + 2y) y' = 3x^2 - y cos(xy)

y' = (3x^2 - y cos(xy)) / (x cos(xy) + 2y)