Example of a function whose directional derivatives are always positive I need an example of a function  $f: \Bbb R^n \rightarrow  \Bbb R$ such that it`s directional derivative at the direction of the vector $y$ is such that $\mathbf{D_y}(a)>0$ for a fixed vector $y$ and every $a$ in $\Bbb R^n$
I'm having my doubts with this one since I just proved that there no $f: \Bbb R^n \rightarrow  \Bbb R$ such that $\mathbf{D_y}(a)>0$ for every  $y$ at a fixed point $a$ in $\Bbb R^n$ by using the property that states that $\mathbf{D_y}(a)=y \nabla f(a)$.
 A: Since $y=(y_1,\ldots,y_n)$ is fixed, if you have $\nabla f(a)=(y_1,\ldots,y_n)$ you will have $y\cdot \nabla f(a)=y_1^2+\cdots+y_n^2>0$ (of course, if $y\ne 0$). 
This can be easily achieved by taking $f(x_1,\ldots,x_n)=y_1x_1+\cdots+y_nx_n$. In shorter notation, you can define $f(x)=y\cdot x$.
A: My understanding of your question is as following:
For each $y \in \mathbb{R}^n$ find a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ such that the directional derivative $\partial_yf = \mathbf{D}_yf$ is positive.
Please note that for $\|y\|=0$ the directional derivative is always $0$, so I'll exclude that case.

I suggest the following approach:
For an arbitrary $y \in \mathbb{R}^n \setminus \lbrace 0 \rbrace$ let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be given by
$$
    f(x) = \sum_{i=1}^n y_i x_i
$$
Then we have for $i \in \lbrace 1, \ldots, n \rbrace$ and $a \in \mathbb{R}^n$
$$
    \frac{\partial f}{\partial x_i}(a) = y_i
$$
which means that it is $\nabla f(a) = y$.
This immediately shows that the directional derivative of $f$ in the direction of $y$ is positive:
$$
    \mathbf{D_y}f(a) = y \cdot \nabla f(a) = y \cdot y = \|y\|^2 > 0
$$
