$\qquad f(tx)=t^2f(x)\iff\left\langle \nabla f(x),x\right\rangle =2f(x)$ $f:\mathbb{R}^n\to\mathbb{R}$ is a differentiable function. How do I show that the following are equivalent:
(i) $\qquad f(tx)=t^2f(x)\quad\ \; \ \qquad \forall t\gt 0\land x\neq 0 $
(ii)$\qquad \left\langle \nabla f(x),x\right\rangle =2f(x) \qquad\forall x\neq 0$
This showed up in a set of practice questions and I don't know how to go about answering the question. Thank you. 

If (i) is true let $\phi(t) = f(tx)$ and so $\phi'(t) = \langle \nabla f (tx), x \rangle = 2 t f(tx)$. Setting $t=1$ gives the desired result.
If (ii) is true let $\eta(t) = {1 \over t^2} f(tx)$ and so
$\eta'(t) = {1 \over t^3} (\langle \nabla f (tx), tx \rangle-f(tx)) = 0$. Hence $\eta(t) = \eta(1)$ and we get the desired result.
