I'll assume you're familiar with the fact that a function is convex if and only if its epigraph is convex.
If the function is positive homogenous, then by just checking definitions, we see that its epigraph is a cone. That is, for all $a > 0$, we have:
$$ \begin{aligned}(\mathbf{x},t) \in \text{epi f} &\Leftrightarrow f(\mathbf{x})\le t \\& \Leftrightarrow af(\mathbf{x})=f(a\mathbf{x}) \le at \\ &\Leftrightarrow (a\mathbf{x},at) \in \text{epi f}\end{aligned}$$
On the other hand suppose the epigraph is a cone. This means for all $a>0$, if $(\mathbf{x},t) \in \text{epi} f$ then $(a\mathbf{x},at) \in \text{epi f}$. Clearly $(\mathbf{x},f(\mathbf{x})) \in \text{epi f}$, so $(a\mathbf{x},af(\mathbf{x})) \in \text{epi f}$, which means that
$$ f(a\mathbf{x}) \le a f(\mathbf{x}).$$
Likewise $(a\mathbf{x},f(a\mathbf{x})) \in \text{epi f}$. So, again, if the epigraph is a cone, $(\mathbf{x},f(a\mathbf{x})/a) \in \text{epi f}$, or:
$$ f(\mathbf{x}) \le f(a\mathbf{x})/a.$$
Combining the two inequalities, we get:
$$ f(\mathbf{x}) \le f(a\mathbf{x})/a \le f(\mathbf{x})$$
Which means $f(a\mathbf{x}) = a f(\mathbf{x})$, ie, the function is positive homogenous.
Best Answer
The union of the 1st and the 3rd quadrants is a cone but not convex; the 1st quadrant itself is a convex cone.