How to find the stationary point of $f(x,y)=\sin x \sin y \sin (x+y)$ How to find the stationary point(s) of $f(x,y)=\sin x \sin y \sin (x+y)$
With $x,y\in(0,\pi)$
So far I have found $$\nabla f =(\color{red}{\sin x\cos (x+y)+\cos x\sin (x+y)\sin y},\color{blue}{\sin y\cos (x+y)+\cos y\sin (x+y)\sin x)}$$
So we need 
$$(\sin x \cos (x+y)+\cos x \sin (x+y)) \sin y=0 \\ (\sin y \cos (x+y)+\cos y\sin (x+y))\sin x=0$$
Problem is I don't know how to solve this.

$\left(\sin x \cos(x+y) + \cos x\sin(x+y)\right) \sin y = 0$ when $\sin x \cos(x+y) + \cos x\sin(x+y) = 0$ or when $\sin y = 0$.  I trust that you know when $\sin y = 0$.  What about that first one?
\begin{align}
  \sin x \cos(x+y) + \cos x\sin(x+y) &= 0 \\
  \sin x \cos(x+y) &= -\cos x\sin(x+y)\\
  -\tan x &= \tan(x+y)\\
  \tan(-x) &= \tan(x+y)
\end{align}
Because tangent is periodic with period $\pi$, then this equation is satisfied when $x+y = -x + n\pi$, where $n$ is any integer (provided there are no domain restriction violations).  This is equivalent to $y = -2x + n\pi$.
The second equation you mentioned is nearly identical.
