Functions with the sum of its partial derivatives zero

Question

I have a multivariable function $G(x_i,x_j,x_k)$ which satisfies $G(x,x,x)=0$. Does it implies that the sum of its partial derivatives vanishes at $(x,x,x)$, i.e., ${\frac{\partial G(x_i,x_j,x_k)}{\partial x_i}}|{(x,x,x)}+{\frac{\partial G(x_i,x_j,x_k)}{\partial x_j}}|{(x,x,x)}+{\frac{\partial G(x_i,x_j,x_k)}{\partial x_k}}|{(x,x,x)}=0$.

I can varify this from some of examples such that $G(x_i,x_j,x_k)=(x_j+x_k-2x_i)$, $G(x_i,x_j,x_k)=x_j^2x_k-x_i^3$.

But I am unable to prove this in general. Can anyone help me to guide through its proof?

Answer 1

It's just the multivariable chain rule, we have $G(x,y,z)$ and $x=y=z=t$, then:

$$ \frac{dG}{dt} = \frac{\partial G }{\partial x} \frac{dx}{dt}+ \frac{\partial G}{\partial y} \frac{dz}{dt}+\frac{\partial G }{\partial z} \frac{dx}{dt}$$

With $ \frac{dx}{dt} = \frac{dy}{dt} = \frac{dz}{dt}=1$, and yes of course, $ \frac{dG}{dt}=0$ since $G$ is constant on this path (line through origin with direction $(1,1,1)$