Given, $u=\log \left(x^3+y^3+z^3-3 x y z\right)$
$$
\begin{aligned}
& u_x=\frac{d u}{d x}=\frac{3 x^2-3 y z}{\left(x^3+y^3+z^3-3 x y z\right)} \\
& u_y=\frac{d u}{d y}=\frac{3 y^2-3 x z}{x^3+y^3+z^3-3 x y z}
\end{aligned}
$$
and $u_z=\frac{d u}{d z}=\frac{3 z^2-3 x y}{x^3+y^3+z^3-3 x y z}$
$$
\begin{aligned}
& u_x+u_y+u_z \\
& \quad=\frac{3\left(x^2+y^2+z^2-x y-y z-z x\right)}{(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)} \\
& \Rightarrow \quad(x+y+z)\left(u_x+u_y+u_z\right)=3
\end{aligned}
$$