$\frac{d y}{d x}=\sin (x-y)+\cos (x-y)$

Put $x-y=v$
$$
\begin{aligned}
& 1-\frac{d y}{d x}=\frac{d v}{d x} \\
& \Rightarrow 1-\frac{d v}{d x}=\sin v+\cos v \\
& \Rightarrow 1-(\sin v+\cos v)=\frac{d v}{d x} \\
& \Rightarrow \frac{d v}{1-(\sin v+\cos v)}=d x \\
& \Rightarrow \frac{d v}{1-\frac{2 \tan \frac{v}{2}}{1+\tan ^2 \frac{v}{2}}-\frac{1-\tan ^2 \frac{v}{2}}{1+\tan ^2 \frac{v}{2}}}=d x \\
& \Rightarrow \frac{\left(1+\tan ^2 \frac{v}{2}\right) d v}{1+\tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}-1+\tan ^2 \frac{v}{2}}=d x \\
& \Rightarrow \frac{\sec ^2 \frac{v}{2} d v}{2 \tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}}=d x \\
&
\end{aligned}
$$

Let $\tan \frac{v}{2}=u \Rightarrow \frac{1}{2} \sec ^2 \frac{v}{2} d v=d u$
$$
\begin{aligned}
& \Rightarrow \frac{d u}{u^2-u}=d x \Rightarrow \frac{d u}{u(u-1)}=d x \\
& \Rightarrow \int\left(\frac{-1}{u}+\frac{1}{u-1}\right) d u=\int d x \\
& \Rightarrow x+C=\log |u-1|-\log |u| \\
& \Rightarrow x+C=\log \left|\tan \frac{v}{2}-1\right|-\log \left|\tan \frac{v}{2}\right|
\end{aligned}
$$
$\Rightarrow x+C=\log \left|\frac{\tan \frac{x-y}{2}-1}{\tan \frac{x-y}{2}}\right|$