Definition: A function $f: \mathbb{R}^n \to \mathbb{R}$ is said to be homogenous of degree $k$ if $\forall t \in \mathbb{R}$ and $(x_1, \dots , x_n) \in \mathbb{R}^n$ the equations $f(tx_1, \dots , tx_n)= t^k f(x_1, \dots , x_n)$ holds
Problem: Show that for $f \in C^1( \mathbb{R}^n \setminus \lbrace 0 \rbrace$) the following equation holds $$\sum_{i=1}^n x_i \frac{\partial f}{\partial x_i}(x_1, \dots ,x_n)= kf(x_1, \dots , x_n)$$
My approach: Following my first impulse, I thought it might be best to use induction and I managed to solve for the base case, let $n=1$ then we have $f: \mathbb{R} \to \mathbb{R}$ a real valued function with property $$f(tx)=t^kf(x) \overset{\displaystyle \frac{d}{dt}}{\implies} f'(tx)x = kt^{k-1}f(x) $$ Now let $t=1$ then we obtain $$ \overset{t=1}{\implies} f'(x) x = \frac{\partial f(x)}{\partial x}x =ktf(x) $$
which would be sufficient for my base case $n=1$, now I struggle with the step $n \leadsto n+1$ $$\sum_{i=1}^{n+1} x_i \frac{\partial f}{\partial x_i}(x_1, \dots , x_n, x_{n+1}) = \dots $$ My idea was to split up the sum, I believe what confuses me the most is the notation, I am unable to continue there. I would appreciate hints or showing me a few additional steps.
Correct approach. Take $g(t)=f(tx)$. Then $$ g'(t)=\frac{d}{dt}f(tx)=\frac{d}{dt}f(tx_1,\ldots,tx_n)=x_1f_{x_1}(tx_1,\ldots,tx_n)+\cdots+ x_nf_{x_n}(tx_1,\ldots,tx_n), $$ and hence $$ g'(1)=x_1f_{x_1}(x_1,\ldots,x_n)+\cdots+x_nf_{x_n}(x_1,\ldots,x_n)=x\cdot\nabla f(x). $$ Also $g(t)=t^kf(x),$ and hence $g'(1)=kf(x)$. Thus $$ x\cdot\nabla f(x)=kf(x). $$
Note that $$ \frac{d}{dt}f\big(g_1(t),\cdots,g_n(t)\big)=g'_1(t)f_{x_1}\big(g_1(t),\cdots,g_n(t)\big)+\cdots+ g'_n(t)f_{x_n}\big(g_1(t),\cdots,g_n(t)\big). $$