Let $\alpha \in \mathbb{R}$ and $f:\mathbb{R^n} \to \mathbb{R}$ be a differentiable function that satisfies $f(tx) = t^\alpha f(x)$ for all $x \in \mathbb{R^n}$ and $t \in (0, \infty)$.
How can one prove that
$$\alpha f(x) = \sum_{k=1}^n \frac{\partial f(x)}{\partial x_k} x_k$$
for all $x \in \mathbb{R}^n$?
It says: Let $\varphi(t) = f(tx)$. Calculate $\varphi'(1)$ in two ways. Can someone tell me how it's done?
I know that by starting with $f(tx)=t^\alpha f(x)$ and differentiating both sides with respect to $t$, we get $$ \sum_{k=1}^n\frac{\partial f(tx)}{\partial x_i}x_i=\alpha t^{\alpha-1}f(x) $$ by the chain rule, and then if we set $t=1$ then the above becomes $$ \sum_{k=1}^n\frac{\partial f(x)}{\partial x_i}x_i=\alpha f(x)$$
But I have no clue on how to calculate $\varphi'(1)$ in two ways.
You managed perfectly. You actually evalueted $\varphi'(1)$ The two methods were the left hand side and the right hand side of your last equation.
Again, the first way to evaluate $\varphi'(1)$ was using it's definition and the chain rule in $\mathbb{R} ^n$
$$ \varphi'(t)= \frac{\partial f (tx)}{\partial t} = \sum_{k=1}^n\frac{\partial f(tx)}{\partial x_i}x_i $$ The second,using the special property of $f$:
$$ \varphi'(t)=\frac{\partial f (tx)}{\partial t}= \frac{\partial t^\alpha f (x)}{\partial t} =\alpha t^{\alpha-1}f(x) $$
Setting now $t=1$ you recover the two ways and this completes the proof.