Answer
$$\frac{{dy}}{{dx}} = - \frac{y}{{x - \frac{1}{y}\cos \left( {\frac{1}{y}} \right) + \sin \left( {\frac{1}{y}} \right)}}$$
Work Step by Step
$$\eqalign{
& y\sin \left( {\frac{1}{y}} \right) = 1 - xy \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {y\sin \left( {\frac{1}{y}} \right)} \right) + \frac{d}{{dx}}\left( {xy} \right) = 0 \cr
& \cr
& {\text{Use the product rule}} \cr
& y\frac{d}{{dx}}\left( {\sin \left( {\frac{1}{y}} \right)} \right) + \sin \left( {\frac{1}{y}} \right)\frac{d}{{dx}}\left( y \right) + x\frac{d}{{dx}}\left( y \right) + y\frac{d}{{dx}}\left( x \right) = 0 \cr
& \cr
& {\text{Find derivatives}} \cr
& y\cos \left( {\frac{1}{y}} \right)\frac{d}{{dx}}\left( {\frac{1}{y}} \right) + \sin \left( {\frac{1}{y}} \right)\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0 \cr
& y\cos \left( {\frac{1}{y}} \right)\left( { - \frac{1}{{{y^2}}}} \right)\frac{{dy}}{{dx}} + \sin \left( {\frac{1}{y}} \right)\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0 \cr
& - \frac{1}{y}\cos \left( {\frac{1}{y}} \right)\frac{{dy}}{{dx}} + \sin \left( {\frac{1}{y}} \right)\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0 \cr
& \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& - \frac{1}{y}\cos \left( {\frac{1}{y}} \right)\frac{{dy}}{{dx}} + \sin \left( {\frac{1}{y}} \right)\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} = - y \cr
& \left( { - \frac{1}{y}\cos \left( {\frac{1}{y}} \right) + \sin \left( {\frac{1}{y}} \right) + x} \right)\frac{{dy}}{{dx}} = - y \cr
& \frac{{dy}}{{dx}} = - \frac{y}{{x - \frac{1}{y}\cos \left( {\frac{1}{y}} \right) + \sin \left( {\frac{1}{y}} \right)}} \cr} $$
