Question: How can I solve $y-xy'-\sin(y')=0$? Are there any general techniques for solving ODE of the form $y=f(y')$, where $f$ is a trigonometric function?

Answer: 
Assuming you mean $y(x)$, what you have is what is known as a [Clairaut's equation](http://en.wikipedia.org/wiki/Clairaut%27s_equation).

Separate your $y$ and $y'$ and then differentiate with respect to $x$, then factor and solve.

After a bit of manipulation, you should get $y = \sin \alpha + \alpha x$ or $y = \pm(x \arccos (-x) + \sqrt{-x^2 +1})$.
May be this helps

$y−xy′−sin(y′)=0 \to sin(y')=y-xy'\\
y'-(xy''+y')-y''cos(y')=0 \to cos(y')=-\frac{xy''}{y''}=-x\\
sin^2(y')+cos^2(y')=1 \to (y-xy')^2+x^2=1 \to y^2+x^2(y')^2-2xyy'+x^2=1$

Which is a first-order nonlinear ordinary differential equation, that must be solvable.

Let $y=xv(x)$.

After doing the math we will get:
$y(x) = x (-sin^{-1}(x)-\frac{\sqrt{-x^2+1}}{x}+c\_1)$ or $y(x) = x (sin^{-1}(x)+\frac{\sqrt{-x^2+1}}{x}+c\_1)$