University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dx}=\frac{y}{-x-\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)}$$
$$y\sin\Big(\frac{1}{y}\Big)=1-xy$$ 1) Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}\Big(y\sin\Big(\frac{1}{y}\Big)\Big)=\frac{d}{dx}(1-xy)$$ $$\frac{dy}{dx}\sin\Big(\frac{1}{y}\Big)+y\cos\Big(\frac{1}{y}\Big)\frac{d}{dx}\Big(\frac{1}{y}\Big)=0-(y+x\frac{dy}{dx})$$ $$\frac{dy}{dx}\sin\Big(\frac{1}{y}\Big)+y\cos\Big(\frac{1}{y}\Big)\Big(-\frac{1}{y^2}\Big)\frac{dy}{dx}=-x\frac{dy}{dx}-y$$ $$\frac{dy}{dx}\sin\Big(\frac{1}{y}\Big)-\frac{1}{y}\cos\Big(\frac{1}{y}\Big)\frac{dy}{dx}=-x\frac{dy}{dx}-y$$ 2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$: $$-x\frac{dy}{dx}-\frac{dy}{dx}\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)\frac{dy}{dx}=y$$ $$\frac{dy}{dx}\Big(-x-\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)\Big)=y$$ $$\frac{dy}{dx}=\frac{y}{-x-\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)}$$