 Solution

Let $L(x,y,z)=(x-y,y-z,z-x)$. This is a linear map, hence smooth; composing it with the given $C^1$ function proves that $g=f\circ L$ is continuously differentiable by the chain rule. Write $f_1,f_2,f_3$ for the first partial derivatives, evaluated at $L(p)$. ThenAdding these identities givesGeometrically, $L(p+t(1,1,1))=L(p)$, so $g$ is constant along every line in the diagonal direction; its derivative in that direction must vanish.