Not that easy 2

$\begin{aligned} u & = \ln (x^3+y^3+z^3-3xyz) & \small \color{#3D99F6} \text{By chain rule,} \\ \frac {\partial u}{\partial x} & = \frac 1{x^3+y^3+z^3-3xyz} \cdot (3x^2 - 3yz) \\ & = \frac {3x^2 - 3yz}{x^3+y^3+z^3-3xyz} & \small \color{#3D99F6} \text{Similarly,} \\ \frac {\partial u}{\partial y} & = \frac {3y^2 - 3zx}{x^3+y^3+z^3-3xyz} \\ \frac {\partial u}{\partial z} & = \frac {3z^2 - 3xy}{x^3+y^3+z^3-3xyz} \end{aligned}$

Therefore,

$\begin{aligned} x \frac {\partial u}{\partial x} + y \frac {\partial u}{\partial y} + z \frac {\partial u}{\partial z} & = \frac {3x^3 - 3xyz + 3y^3 - 3xyz + 3z^3 - 3xyz}{x^3+y^3+z^3-3xyz} \\ & = \frac {3(x^3+y^3+z^3-3xyz)}{x^3+y^3+z^3-3xyz} \\ & = \boxed{3} \end{aligned}$

Given that, $u= ln(x^3 + y^3+z^3-3xyz)$ ... Our aim is to find out partial derivative of the function $f(x,y,z)=u$ .. So we can proceed on that way.... \Rightarrow x\frac{\delta u}{\delta x} +y\frac{\delta u}{\delta y}+ z\frac{\delta u}{\delta z} = \frac{3(x^3+y^3+z^3-3xyz)}{(x^3+y^3+z^3-3xyz)}=\boxed{\color\red{3}}