Consider a monotonically increasing and differentiable function $y=f(x)$ that passes through the origin.
$\gamma=\{(x,y)|y=f(x)\}$ is the graph of $f$.
Claim: there exists a ray $R$ such that $R\cap \gamma= \{p\}$ and $f'(p)=R'$,
That is, the ray $R$ touch $\gamma$ at exactly one point $p$ and L is a tangent for $\gamma$ and $p$ is not the endpoint of the ray.
Intuitively the claim is correct. Here is my attempts to prove it.
Since $f$ is increasing and passes through the origin, there must exist two lines $L_1: y=a_1x$ and $L_2: y=a_2x$ such that $\gamma$ is bounded inside of the region between $L_1$ and $L_2$. $a_1,a_2\in[0,+\infty]$
Suppose $a_1>a_2$ (WLOG). Let $q$ be one the points that $\gamma$ and $L_2$ touch. By the differentiability of $f$, $\exists \epsilon>0$ such that $f$ is locally concave inside the ball $B(q,\epsilon)$.
(somethings might be missing here)
Case 1: there exists a point $r$ on $\gamma$ such that $r$ is also on $L_2$ and $r_x>q_x$. Then when we are moving a point on $\gamma$ from $q$ to $r$, there must exist at least one connected and non-singleton set $(P_1, P_2, ...)$ of points on $\gamma$ such that $f'(p_i)> a_1$ $\forall p_i\in P_i\forall i$; Let $P_1$ be the first set of this kind, with limiting end points $p_1$ and $q_1$. By definition, $f'(p_1)\leq a_1$. So, between $p_1$ and $q_1$ there must exist a point $p_1^*\neq p$ such that $f$ is concave between $p_1$ and $p_1^*$. Picking any point inside of $(p_1,p_1^*)$ on $\gamma$ we can construct an eligible ray $R$.
Case 2: there does not exist a point $r$ on $\gamma$ such that $r$ is also on $L_2$ and $r_x>q_x$. Then there must exist $r$ on $\gamma$ such that $r$ is also on $L_2$ and $r_x<q_x$. The rest of the proof is similar to the Case 1.
Is my attempt seem worthy? Seems overly complicated for such a simple fact, although the proof only works for a subset of monotone functions.