Answer: $\frac{5}{\sqrt[5]{256}}$.
Let $\frac{x^2}{y} = a$, $\frac{y^2}{z} = b$ and $\frac{z^2}{x} = c$. The problem conditions in this new variables take the following form
---

$$abc = 1, \quad a+b+c = ab+bc+ca.$$
Now we readily get $(a-1)(b-1)(c-1) = 0$. Therefore, at least one of the numbers $a, b, c$ should be equal to $1$. Without loss of generality we assume that $a=1$. Then $y = x^2$ and $z = \frac{1}{x^3}$.
If $0 < x \le 1$ then $x^2 \le x \le \frac{1}{x^3}$ and if $x > 1$ then $\frac{1}{x^3} < x < x^2$. Hence in all cases $x$ is a median: $t_2 = x$. Finally by AM-GM inequality we get
$$\frac{x^2 + \frac{1}{x^3}}{x} = x + \frac{1}{x^4} = \frac{x}{4} + \frac{x}{4} + \frac{x}{4} + \frac{x}{4} + \frac{1}{x^4} \ge 5\sqrt[5]{\frac{1}{4^4}}.$$
The equality holds when $\frac{x}{4} = \frac{1}{x^4}$ or $x = \sqrt[5]{4}$. In this case $y = x^2 = \sqrt[5]{16}$ and $z = \frac{1}{x^3} = \frac{1}{\sqrt[5]{64}}$. Thus, $\frac{t_1+t_3}{t_2}$ takes its smallest value $\frac{5}{\sqrt[5]{256}}$ at $x = \sqrt[5]{4}$, $y = \sqrt[5]{16}$, and $z = \frac{1}{\sqrt[5]{64}}$. Done.