Step 1:
Given :$(x-y)(dx+dy)=dx-dy$
On expanding and rearranging we get,
$x(dx-dy)-y(dx-dy)=dx-dy$
$(x-y+1)dy=(1-x+y)dx$
$\large\frac{dy}{dx}=\frac{(1-x+y)}{(x-y+1)}$
Step 2:
Let $(x-y)=t$ then $1-\large\frac{dy}{dx}=\frac{dt}{dx}$ or $\large\frac{dy}{dx}=$$1-\large\frac{dt}{dx}$
$1-\large\frac{dt}{dx}=\frac{(1-t)}{(1+t)}$
$-\large\frac{dt}{dx}=\frac{(1-t)}{(1+t)}$$-1$
$-\large\frac{dt}{dx}=\frac{-2t}{(1+t)}$
$\large\frac{dt}{dx}=\frac{2t}{(1+t)}$
Separating the variables
$\large\frac{(1+t)dt}{2t}$$=dx$
$\large\frac{dt}{2t}+\frac{dt}{2}$$=dt$
Step 3:
Integrating on both sides we get,
$\large\frac{1}{2}$$\log t+\large\frac{1}{2}$$t=t+C$
substituting for t we get,
$\large\frac{1}{2}$$\log (x-y)+\large\frac{1}{2}$$(x-y)=(x-y)+C$
$\large\frac{1}{2}$$\log (x-y)=\large\frac{(x+y)}{2}$$+C$
$\log(x-y)=(x+y)+2C$
Step 4:
To evaluate for $C$ let us substitute the given values of $x=0$ and $y=-1$
$\log(1)=-1+k$(where $2C=k$)
$k=1$
$\log(x-y)=(x+y)+1$
$\log|x-y|=x+y+1$
This is the required solution.