There are five critical points. Here's why:
f(x)=-xsin(x); f'(x)=-xcos(x)-sin(x)=0 at a critical point.
sin(x)=-xcos(x); tan(x)=-x. x=0 is a solution so there is a critical point at (0,f(0))=(0,0).
f''(x)=xsin(x)-2cos(x); at x=0, f''(x)<0 so (0,0) is a maximum.
Since the origin is a maximum we know that the curve is decreasing on each side, so f(x) is becoming more negative. We also know that f(x)=0 when x=π and x=-π as well as at the origin. Therefore, there must be a turning point (minimum) between x=0 and x=±π. There are two minima on either side of the origin.
Between π and 2π and between -π and -2π there must be maxima because, when -2π<x<-π, sin(x)>0 and -x>0, therefore f(x)>0. When π<x<2π, -x<0 and sin(x)<0, therefore f(x)>0. To go from 0 back to zero (sin(±2π)=0) there has to be a maximum. 2π=6.28 approx, which is a little more than 6, so the curve has already passed the maximum. In all we have five critical points: x=0, x between 0 and π (min), x between π and 2π (max) and, by symmetry the same for negative x up to x=-6. That's one max in the origin and a pair of extrema on each side of the origin.