$1 + (\frac{x}{y} - \sin(y))y' = 0$
Rewrite the equation, we have: $1dx + (\frac{x}{y} - \sin(y))dy = 0$
Let $M = 1$, and $N = \frac{x}{y}- \sin(y)$
Then $M_y = 0$, and $N_x= \frac{1}{y}$, where $M_y \neq N_x$
So, it is not exact.
Let $R_1 = \frac{M_y - N_x}{M} = \frac{-1}{y}$
So $\mu = e^{-\int R_1 dy} = e^{\int \frac{1}{y} dy} = e^{\ln{y}} = y$
Then multiply $\mu = y$ on both sides, we have
$y dx + (x - y\sin(y)) dy = 0$
and now it is exact since $M_y = 1 = N_x$
So $\exists\ \psi(x,y)$, such that $\psi_x = M = y$
$\psi = \int M dx = \int y dx = yx + h(y)$
Then, we can get $\psi_y = x + h'(y)$
Also, we know $\psi_y = N = x - y\sin(y)$, which implies that
$h'(y) = - y\sin(y)$
So $h(y) = \int h'(y) dy = \int - y\sin(y) dy$
By integrating by parts, we can know
$ h(y) = y\cos{y} -\sin{y} + c$
Therefore, we can have
$\psi = yx + y\cos{y} -\sin{y} = c$
