A symmetric partial differential equation

Question

Is there a function $f(x,y,z)$ such that all its partial derivatives have the same form $f_x=f_y=f_z=\frac{1}{xyz}$?

Answer 1

Suppose that a function $f(x,y,z)$ exists such as $f_x=f_y=\frac{1}{xyz}$

$\frac{\partial f_x}{\partial y} =f_{xy}=\frac{\partial \frac{1}{xyz}}{\partial y }= -\frac{1}{xy^2z}$

$\frac{\partial f_y}{\partial x}=f_{yx}=\frac{\partial \frac{1}{xyz}}{\partial x }= -\frac{1}{x^2yz}$

Since $f_{xy}=f_{yx}$ this would implies $-\frac{1}{xy^2z}=-\frac{1}{x^2yz}$

This is possible only if $x=y$, but not any $x$ and any $y$. Thus the supposition is false.

As a consequence, such function doesn't exist.

A-fortiori for the three variables function.