The difference between the greatest and least values of the function $f(x)=\cos x+\frac{1}{2} \cos 2 x-\frac{1}{3} \cos 3 x$, is 

$9 / 4$

Clearly, f(x) is a periodic function with period $2 \pi$. So, the difference between the greatest and the least values of the function f(x) is the difference between the greatest and the least values on the interval $[0,2 \pi]$.

Let us find these values on the interval $[0,2 \pi]$. We have,

$f'(x)=-(\sin x+\sin 2 x-\sin 3 x)$

$\Rightarrow \quad f^{\prime}(x)=-\left(2 \sin \frac{3 x}{2} \cos \frac{x}{2}-2 \sin \frac{3 x}{2} \cos \frac{3 x}{2}\right)$

$\Rightarrow \quad f^{\prime}(x)=-2 \sin \frac{3 x}{2}\left(\cos \frac{x}{2}-\cos \frac{3 x}{2}\right)$

$\Rightarrow \quad f^{\prime}(x)=-4 \sin \frac{x}{2} \sin \frac{3 x}{2} \sin x$

∴  $f^{\prime}(x)=0 \Rightarrow x=0,2 \pi / 3, \pi$ and $2 \pi$

Now,

$f(0)=1+\frac{1}{2}-\frac{1}{3}=\frac{7}{6}, f\left(\frac{2 \pi}{3}\right)=-\frac{13}{12}, f(\pi)=-\frac{1}{6}$

and, $f(2 \pi)=\frac{7}{6}$

The largest and the smallest of these values are 7/6 and -13/12 respectively.

Hence, the required difference = 7/6 - (-13/12) = 9/4.