$\text{ If }x+y=\sin (x+y)\text{ then }\frac{dy}{dx}=$

$\begin{array}{r}\text{ Given }x+y=\sin (x+y)\\ \text{ Differentiate both sides }\end{array}$

$\begin{array}{c}1+\frac{dy}{dx}=\cos \left(x+y\right)\frac{d}{dx}\left(x+y\right)\\ 1+\frac{dy}{dx}=\cos \left(x+y\right)\left(1+\frac{dy}{dx}\right)\\ 1+\frac{dy}{dx}-\cos \left(x+y\right)\left(1+\frac{dy}{dx}\right)=0\\ \left(1+\frac{dy}{dx}\right)\left(1-\cos \left(x+y\right)\right)=0\end{array}$

$\text{ It implies }1+\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-1$