Maximum value of $f(x, y, z) = e^{xyz}$ in the domain x+y+z = 3. Need the maximum value of f$(x, y, z) = e^{xyz}$ in the domain $x+y+z = 3$.

Please help me on this, as I could not find how to solve this one. 

Thanks in advance.
Assuming $x,y,z \ge 0$

$e^{xyz}$ is maximum if $xyz$ is maximum

$AM \ge GM\\
\implies \dfrac{x+y+z}{3} \ge\sqrt[3]{xyz}\\
\implies \sqrt[3]{xyz} \le 1\\
\implies xyz \le 1$

Maximum value of $e^{xyz}=e^{1}=e$
For e^xyz the domain is all (x,y,z) ,so the domain is whole space