Answer
$\frac{dy}{dx}=\frac{1-\cos(x+y)}{\cos (x+y)+\sin y}$
Work Step by Step
Implicitly differentiate $\sin (x+y)=x+\cos y$ with respect to $x$:
$[\cos (x+y)](1+\frac{dy}{dx})=1+(-\sin y\times \frac{dy}{dx})$
Or
$\cos(x+y)+\cos (x+y)\frac{dy}{dx}=1-\sin y \times\frac{dy}{dx}$
$\implies cos(x+y)\frac{dy}{dx}+\sin y\frac{dy}{dx}$$=1-\cos (x+y)$
Or $\frac{dy}{dx}=\frac{1-\cos(x+y)}{\cos (x+y)+\sin y}$
