The curve, which satisfies the differential equation $\frac{xdy-ydx}{xdy+ydx}= y^2 sin (xy)$ and passes through (0, 1), is given by

y (1 - cos xy) + x = 0

The given differential equation can be written as

$(\frac{xdy-ydx}{x^2})(\frac{x^2}{y^2})= (x\, dy + y\, dx) sin\, xy$

or $d(\frac{y}{x})(\frac{x^2}{y^2}) = d (xy) sin\, xy$

Integrating both the sides we get,

$-\frac{1}{y/x} = - cos\, xy + c ⇒ = cos (xy) - c$

For x = 0, y = 1, we get c = 1.

Hence (A) is the correct answer.