Let $f:\mathbb R^n\to \mathbb R$ be a function which satisfies $$\underline{x}\cdot(\nabla f(\underline{x}))\ge 0\,\,\forall\underline{x}\in\mathbb R^n.$$ Fix $\underline{x}\in\mathbb R^n$ and define $g:\mathbb R\to\mathbb R$ by $g(t)=f(t\underline{x})$.
a) Prove that $g'(t)\ge 0\,\,\forall t>0$.
b) Deduce that $f$ has a minimum at $\underline{0}$.
For part (a) I got: $$x_1\frac{\partial(f(tx_1,...,tx_n))}{\partial(x_1t)}+...+x_n\frac{\partial(f(tx_1,...,tx_n))}{\partial(x_nt)}$$
Then I'm not sure what to do. Maybe let $t=1$ and use the information at the top? But it says for all $t>0$...
Part (b) I have no idea.
For $(a)$ there is no need to evaluate at any particular point. You need to notice that $$tg'(t) = (t\textbf x)\cdot \nabla f(t\textbf x) \ge 0.$$ Since $t > 0$, then also $g'(t) \ge 0$.
For $(b)$, suppose that there is $\textbf x \in \mathbb{R}^n$ such that $f(\textbf x) < f(\textbf 0)$ and consider the $g$ function defined as above, relative to this particular $\textbf x$. Then the Mean Value Theorem gives a contradiction, indeed $$0 > f(\textbf x) - f(\textbf 0) = g(1) - g(0) = g'(\theta) \ge 0,$$ where $\theta \in (0,1)$.