$Let \ f=\sin x \sin y \sin (x+y)$ Differentiate f partially w.r.t x $f_x = \cos x\sin y \sin (x+y)+ \sin x \sin y \cos (x+y)$ Differentiate f partially w.r.t y $f_y= \sin x \cos y \sin (x+y) + \sin x \sin y \cos (x+y)$ To get stationary values, we solve fx=0 and fy=0 simultaneously. $f_x=0 = \cos x \sin y \sin (x+y) + \sin x \sin y \cos (x+y) $ $f_y=0=\sin x \cos y \sin (x+y) + \sin x \sin y \cos (x+y)$ From the above 2 equations, $\cos x \sin y \sin (x+y) = \sin x \cos y \sin (x+y) \ and \ \sin x \sin y \cos (x+ y) =0$ $considering \ \cos x \sin y \sin (x+y) = \sin x \cos y \sin (x+y),$ $\sin(x+y) (\cos x \sin y - \sin x \cos y)=0 $ $\sin (x+y) \sin (x-y)=0$ $x+y=n \pi, \ x-y=n\pi, \ \ n\in I$ $gives \ x=n\pi , \ y=0 $ $ f=\sin n\pi \sin 0 \sin n\pi=0$ $Also, \ f_x=0 = \cos x \sin y \sin (x+y) + \sin x \sin y \cos (x+y)$ $gives, \ \cos x \sin (x+y) + \sin x \cos (x+y) =0$ $So, \ \sin (2x+y)=0 \ \cdots (C)$ Similarly, $from \ f_y=0 = \sin x \cos y \sin (x+y) + \sin x \sin y \cos (x+y)$ $\sin (x+2y)=0 \ \cdots (D)$ From (C) and (D), $2x+y=n\pi \ \cdots (E)$ $x+2y=n\pi$ $-2x-4y=-2n\pi \ \cdots (F)$ Adding (E) and (F), $-3y= - n \pi$ $y=-\dfrac {n\pi}{3}$ $x= n\pi + \dfrac {2n\pi}{3} = \dfrac {5n \pi}{3}$ $f=\sin \left (- \dfrac {n\pi}{3} \right ) \sin \left ( \dfrac {5 n \pi}{3} \right )\sin \left ( \dfrac {4n \pi}{3} \right ) = \pm \dfrac {3\sqrt{3}}{8}$ Hence the stationary values of $\sin x \sin y \sin (x+y) \ are \ 0, \pm \dfrac {3\sqrt{3}}{8}.$