Answer
$\dfrac{dy}{dx}=\dfrac{\cos{x}-\tan{y}-1}{x\sec^2{y}}.$
Work Step by Step
$\sin{x}=x+x\tan{y}\rightarrow$
$\dfrac{d}{dx}(\sin{x})=\dfrac{d}{dx}(x)+\dfrac{d}{dx}(x\tan{y})\rightarrow$
$\cos{x}=1+(\tan{y}+\dfrac{dy}{dx}(x\sec^2{y}))\rightarrow$
$\cos{x}-\tan{y}-1=\dfrac{dy}{dx}(x\sec^2{y})\rightarrow$
$\dfrac{dy}{dx}=\dfrac{\cos{x}-\tan{y}-1}{x\sec^2{y}}.$
