The solution of the differential equation $(1+xy) x\, dy + (1-xy)y\, dx = 0, $ is

$-\frac{1}{xy}+log \left(\frac{y}{x}\right) = C$

The correct answer is option (3) : $-\frac{1}{xy}+log \left(\frac{y}{x}\right) = C$

The given equation can be written as

$(x\, dy + y\, dx) + xy (x\, dy -y\, dx)=0$

$⇒d(xy) +xy (x\, dy -y\, dx)=0$

$⇒\frac{d(xy)}{(xy)^2}+\frac{x\, dy-y\, dx}{xy}= 0$

$⇒\frac{d(xy)}{(xy)^2}+\frac{dy}{y}-\frac{dx}{x}=0$

$⇒\frac{d(xy)}{(xy)^2}+d(log\, y) - d(log\, x)=0$

$⇒\frac{d(xy)}{(xy)^2}+d(\log\, y - log\, x) =0$

$⇒\frac{d(xy)}{(xy)^2}+d\, log \left(\frac{y}{x}\right) = 0 $

On integrating, we get

$-\frac{1}{xy}+ log \left(\frac{y}{x}\right) = C$ as the required solution.