Answer
$$\frac{dy}{dx}=\frac{y}{-x-\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)}$$
Work Step by Step
$$y\sin\Big(\frac{1}{y}\Big)=1-xy$$
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 $dy/dx$:
$$-x\frac{dy}{dx}-\frac{dy}{dx}\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)\frac{dy}{dx}=y$$
$$\frac{dy}{dx}\Big(-x-\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)\Big)=y$$
$$\frac{dy}{dx}=\frac{y}{-x-\sin\Big(\frac{1}{y}\Big)+\frac{1}{y}\cos\Big(\frac{1}{y}\Big)}$$
