$1+(\frac{x}{y} - sin(y))y' = 0$
Let $M(x,y)=1$ and $N(x,y)=(\frac{x}{y} - sin(y))$
Then, $M_y = 0$ and $N_x = \frac{1}{y}$
We can see that this equation is not exact.
However, we know $\frac{N_x-M_y}{M}=\frac{1}{y}$, and $\mu =y$
Multiplying the original equation by $\mu(y)$, we have
$y+(x-ysin(y))y'=0$
Since $M_y=1$ and $N_x=1$, this equation is exact.
Thus, there exists a function $\psi(x,y)$ such that
$\psi_x(x,y)=y$ and $\psi_y(x,y)=x-ysin(y)$
Hence, $\psi(x,y) = \int y dx = xy + h(y)$
then $\psi_y(x,y)= x+h'(y) = x-ysin(y)$
We have $h'(y)=-ysin(y) \Rightarrow h(y)=- \int ysin(y)dy = ycos(y)-sin(y)$
Therefore, $\psi(x,y)=xy+ycos(y)-sin(y)$, and the solutions of the equation are given implicitly by $xy+ycos(y)-sin(y)=C$
