#### Answer

$\frac{dy}{dx} = \frac{-\sin y -y\cos x}{x\cos y + \sin x}$

#### Work Step by Step

$x\sin y + y\sin x = 1$
Differentiate both sides with respect to x
$\sin y + \frac{dy}{dx}x\cos y + \frac{dy}{dx}\sin x + y \cos x = 0$
Isolate $\frac{dy}{dx}$
$\frac{dy}{dx}x\cos y+ \frac{dy}{dx}\sin x = -\sin y - y\cos x$
$\frac{dy}{dx} (x\cos y + \sin x) = (-\sin y - y\cos x)$
$\frac{dy}{dx} = \frac{-\sin y -y\cos x}{x\cos y + \sin x}$