The quotient of two metrics may not be a metric I'm looking for a counterexample to show that $\frac{\rho_1}{\rho_2}$ is not necessarily a metric where $\rho_1$ and $\rho_2$ are metrics. 
We define $\frac{\rho_1}{\rho_2}$ as follows:
$$\frac{\rho_1}{\rho_2}(x,y)=
\begin{cases}\frac{\rho_1(x,y)}{\rho_2(x,y)}, &\text{if }x \neq y \\
0, &\text{if }x=y.
\end{cases}$$
Clearly the triangle inequality should be the targeted.
How should I think to construct the example?

As suggested, take the discrete metric on the top (that which is $1$ whenever the two arguments are unequal and zero if they are equal). Let us consider the space $\mathbb{R^2}$ and take $\rho_2$ to be the Euclidean metric. Then for $x\neq y\in \mathbb{R^2}$, denote $d\equiv \frac{\rho_1}{\rho_2}. $ Then 
$$d(x,y) = \frac{1}{\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}}. $$
Define the points $a = (1,0)$, $b=(4,1)$, $c=(0,1)$, then
$$ d(a,c) = 1/\sqrt{2} \geq d(a,b)+d(b,c) = \frac{1}{4}+\frac{1}{\sqrt{10}} $$
to see that the inequality is true, simply note that $(\frac{1}{4}+\frac{1}{\sqrt{10}})^2<\frac{49}{144}<0.6^2<1/2$. Hence $d$ does not satisfy the metric triangle inequality.
Thus $d$ defined above is a ratio of two metrics but it is not a metric. QED
