The solution of the differential equation $(1+x y) x d y+(1-x y) y d x=0$, is

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

The given equation can be written as

$(x d y+y d x)+x y(x d y-y d x)=0$

$\Rightarrow d(x y)+x y(x d y-y d x)=0 $

$\Rightarrow \frac{d(x y)}{(x y)^2}+\frac{x d y-y d x}{x y}=0$

$\Rightarrow \frac{d(x y)}{(x y)^2}+\frac{d y}{y}-\frac{d x}{x}=0$

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

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

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

On integrating, we get $-\frac{1}{x y}+\log \left(\frac{y}{x}\right)=C$ as the required solution.