$$
1+(\frac{x}{y}-\sin(y))y^{\prime}=0.
$$
\noindent Let
$$
M(x,y)=1\quad \text{and}\quad N(x,y) =(\frac{x}{y}-\sin(y))
$$
Then
$$
\frac{\partial}{\partial y}M(x,y)=0\quad\text{and}\quad \frac{\partial}{\partial x}N(x,y)=\frac{1}{y}
$$
We can see that this equation is not exact, however, note that
$$
\frac{d\mu}{d y}=(\frac{N_x-M_y}{M})\mu=\frac{\mu}{y}\quad\rightarrow\quad \mu=y
$$
Multiplying our original euqation by$\mu(y)$, we have
$$
y+(x-y\sin(y))y^{\prime}=0
$$
We can see that this equaion is exact, since
$$
\frac{\partial}{\partial y}(y)=1=\frac{\partial}{\partial x}(x-y\sin(y))
$$
Thus, there exists a function $\psi(x,y)$ such that
\begin{align}
\psi_x (x,y)&=y\\
\psi_y (x,y)&=x-y\sin(y)
\end{align}
Integating (1) with respect to x, we get
$$
\psi(x,y)=xy+h(y)
$$
for some arbitary function $h$ of $y$. Next, differentiating with respect to $y$, and equating with (2), we get
$$
\psi_y(x,y)=x+h^{\prime}(y)
$$
Therefore,
$$
h^{\prime}(y)=-y\sin(y)\quad\rightarrow\quad h(y)=s\int y\sin(y)dy=y\cos(y)-\sin(y)
$$
and we have
$$
\psi(x,y)=xy+y\cos(y)-\sin(y)
$$
Thus, the solutions of the differential equation are given implicity by
$$
xy+y\cos(y)-\sin(y)=C
$$
