If $f$ continuous differentiable and $f'(r) < 1,$ then $x'=f(x/t)$ has no other solution tangent at zero to $\phi(t)=rt$ Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous differentiable function such that $f(r)=r,$ for some $r.$ Then how to show that 

If $f'(r) < 1,$  then the problem 
  $$x'=f(x/t)$$ has no other solution tangent at zero to $\phi(t)=rt, t>0$.

Tangent here means
$$\lim_{t\to 0^{+}}\frac{\psi(t)-\phi(t)}{t}=0$$
I could only prove that $\psi(0^+)=0,$ and $\psi'(0^+)=r.$ The problem was to use the fact that $f'(r) < 1.$

Peter Tamaroff gave a very good hint in comments. Here is what comes out of it.
Suppose that $x$ is a solution tangent to $rt$ and not equal to it. Since solution curves do not cross, either (i) $x(t)>rt$ for all $t>0$, or (ii) $x(t)<rt$ for all $t>0$. I will consider (i), the other case being similar. 
By assumption, $x/t\to r$ as $t\searrow 0$. From $$f(x/t)=r+f'(r)(x/t-r)+o(x/t-r)$$ and $f'(r)<1$ we obtain
$$t(x/t)'=f(x/t)-x/t = (f'(r)-1)(x/t-r)+o(x/t-r)$$
which is negative for small $t$. This means that $x/t$ increases as $t\searrow 0$, contradicting $x/t\to r$.
