Question: I have to prove that $$x-\frac{x^{3}}{6} < \sin(x) \quad\text{ for }\quad 0<x<\pi $$

I tried to define $f(x) = \sin(x) - (x-\frac{x^{3}}{6})$, and to differentiate it,
but $f'(x) = \cos(x) -1 +\frac{1}{2}x^{2}$. I have no idea how to continue.

Thanks.

Answer: 
In fact $x-x^3/6< \sin x$ for all $x > 0.$ Proof: As you noted, we can define $f(x) = \sin x -x + x^3/6.$ Is $f(x)>0$ for $x > 0?$ Because $f(0)=0,$ this is true if $f'(x) = \cos x -1 + x^2/2 > 0$ for $x>0.$ Is this true? Because $f'(0)=0,$ this will hold if $f''(x) = -\sin x + x > 0$ for $x> 0.$ Well, hopefully the last inequality looks familiar.
Using the approach you've started with, let $f(x)$ be the difference 
$$ f(x) = \sin(x) - (x-\frac{x^3}{6})$$
then we just want to show that this difference is positive for all values $0 < x < \pi$.
This can be done by showing that $f(0) = 0$ and $f'(x)>0$ on the interval.
$$
f'(x) = \cos(x) +\frac{x^2}{2} - 1
$$
The only critical point (where $f'(x)=0$) is at $x=0$, and so you just have to show that $f(x)>0$ for some value on the interval, i.e., $f(1)=.0403$.