$x_i \frac{\partial f(x)}{\partial x_i} = f(x)g_i(x)$ for all $i$

Question

What can we say about the function $f:\mathbb{R}^k \rightarrow \mathbb{R}$ if for all $i=1,...,k$ and all $x \in \mathbb{R}^k$ we have

$x_i\frac{\partial f(x)}{\partial x_i} = f(x)g_i(x)$

Is there actually any restriction we can make other than that $f$ is differentiable?

Answer 1

If $g_i(x)=-1$ then you have that

$\frac{\partial}{\partial x_i}(x_i f(x))=0$

So a solution is

$f(x)=\frac{1}{\prod_{k=1}^n x_k}$

Answer 2

We can say nearly nothing. Let $f(x)$ never equals to zero. Then we can take $g_i(x)=\frac{x_i\frac{\partial f(x)}{\partial x_i}}{f(x)}$

Answer 3

Consider $\phi=\ln|f|$, then $$\frac{∂\phi(x)}{∂x_i}=\frac{g_i(x)}{x_i}.$$ By the Schwarz lemma, the mixed second derivatives do not depend on the order of derivation, thus $$ \frac1{x_i}\frac{∂g_i(x)}{∂x_j}=\frac1{x_j}\frac{∂g_j(x)}{∂x_i} $$ is a necessary condition for $f$ to exist.