Answer
$$\frac{dy}{dx}=\frac{\cos(x+y)}{1-\cos(x+y)}, \quad \cos(x+y)\neq 1.$$
Work Step by Step
By differentiating the equation $ y=\sin (x+y)$ with respect to $ x $, we get
$$\frac{dy}{dx}=(1+\frac{dy}{dx})\cos(x+y)\\
\Longrightarrow \frac{dy}{dx}(1-\cos(x+y))=\cos(x+y)$$
and hence
$$\frac{dy}{dx}=\frac{\cos(x+y)}{1-\cos(x+y)}, \quad \cos(x+y)\neq 1.$$
