Interesting question, althought I am not exactly sure what kind of answer you want. If this answer is not fully satisfactory, please leave a comment.
Why finding the potential function solves the differential equation. Suppose we have a potential function $\phi$ for $(M(x,y),N(x,y))$. Then all the curves $\phi(x,y)=c$ with $c$ a constant form a solution of the ODE. Indeed, if we take the total derivative of $\phi(x,y)=c$ we get
$$
\phi_x(x,y) dx + \phi_y(x,y) dy = 0, \\
M(x,y) dx + N(x,y) dy = 0.
$$
This is a rather formal explanation.
The physics picture. The differential $Mdx + N dy$ can indeed be regarded as the infinitesimal amount of work done by a field $\vec F=(M(x,y),N(x,y))$. This picture can help you understand intuitively why $F(x,y)=c$ solves the ODE $Mdx + Ndy =0$.
Note that a potential in physics is a scalar function $\phi$ such that $-\nabla \phi =\vec{F}=(M,N)$; one adds a minus sign. If such a potential exists, we say that the vector field is exact or, more commonly in physics, conservative. Well-known examples of conservative vector fields are the gravitational force field and electric field in electrostatics.
Conservative vector fields are characterized by the fact that the line integral of them only depend on the end points. Physically this means that the work done by the fields does not depend on the path, only on the beginning and end position. So if $(M(x,y),N(x,y))=\vec{F} = -\nabla \phi$, then
$$
\int_C M(x,y)dx + N(x,y)dy = \int_C \vec{F} = -(\phi(p_2) - \phi(p_1))
$$
where $C$ is a curve going from $p_1$ to $p_2$. This line integral can be interpreted as the work done by the field $\vec F$ along the path $C$. This line integral (i.e. the total amount of work) is zero iff the potential of the endpoints are the same.
Back to the ODE.
Now let us look back at the ODE $M(x,y)dx + N(x,y)dy = 0$. Physically this equation means that the infinitesimal work should be zero. The solution curves $\phi(x,y)=c$ are the equipotential curves. These curves are exactly the paths along which the work is zero for every infinitesimal step.
Best Answer
$\displaystyle 1 + (\frac{x}{y} - \sin y) y'= 0$.
Actually integrating factor that you have is $u(y)$ (you can try with both, first with $u(x)$ and if that doesn't work, with $u(y)$)
$\displaystyle u(y) + u(y)(\frac{x}{y} - \sin y) y'= 0$
$\displaystyle P = u(y), Q = u(y)(\frac{x}{y} - \sin y)$
$\displaystyle P_y = \frac{\partial u}{\partial y}, Q_x = \frac{u(y)}{y}$
To make ODE exact, $P_y = Q_x$ so $u(y) = y$ is a possible integrating factor.