Let $X=R$ and $d_1$ be the normal norm induced metric, and $d_2=\frac{d_1}{d_1+1}$. I want to show that $d_1$ and $d_2$ determine the same topology.
Now, I know that the metric topologies induced by each of $d_1$ and $d_2$, have basis consisting of open balls in $\mathbb{R}$. So if I want to show that the metrics determine the same topology, I then want to show $B_{d_1}(x_1,r_1)\subseteq B_{d_2}(x_2,r_2)$ and $B_{d_2}(x_2,r_2)\subseteq B_{d_1}(x_1,r_1)$ for some $x_1, x_2\in \mathbb{R}$ and $r_1, r_2>0$
First I pick $y\in\mathbb{R}, \epsilon>0$ and consider
\begin{align*} B_{d_1}(y,\epsilon) &= \{z\ |\ d_1(y,z)<\epsilon\} \\ &\supseteq \{z\ |\ d_1(y,x)+d_1(x,z)<\epsilon\} \\ &= \{z\ |\ d_1(x,z)<\epsilon-d_1(y,x)\} \\ &= \{z\ |\ \frac{d_1(x,z)}{d_1(x,z)+1}<\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1}\} \\ &= \{z\ |\ d_2(x,z)<\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1}\}\\ &= B_{d_2}(x,\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1}) \end{align*}
So we have one of the inclusions. Now I'm not sure if this is even allowed? My doubts come from the fact the the ball $B_{d_2}(x,\frac{\epsilon-d_1(y,x)}{d_1(x,z)+1})$ have radius dependent of $z$ which is not fixed.