We can write $Y$ and $Z$ as functions of $X \sim \mathcal U (0, 1)$, i.e. $Y := Y(X)$ and $Z := Z(X)$. To show the expectation is infinite, all we have to do is show $\int_0 ^ 1 \frac{Y(x)}{Z(x)} \ dx = \infty$ (by the Law of the Unconscious Statistician). As you noted, we can write $Z = \min(X, 1 - X)$ and $Y = \max(X, 1 - X)$. But $$
\int_0 ^ {1/2} \frac{Y(x)}{Z(x)} \ dx = \int_0 ^ {1/2} \frac{\max(x, 1 - x)}{\min(x, 1 -x)} \ dx = \int_0 ^ {1/2} \frac{1 - x}{x} \ dx = \infty
$$
One gets the same thing on the interval $[1/2, 1]$ as well after a substitution, so the integral on $[0, 1]$ is $\infty$.
EDIT: Okay, we can get the pdf as well. Let $T(X) = \frac{\max(X, 1 - X)}{\min(X, 1 - X)}$. We calculate $P(T(X) \le t)$ for some $t$. If $t < 1$ then trivially we get $0$, otherwise $$
P(T(X) \le t) = P\left(\frac{1 - X}{X} \le t, X \in (0, 1/2) \right) + P\left( \frac{X}{1 - X} \le t, X \in (1/2, 1)\right).
$$
Careful reflection reveals that the two probabilities on the RHS are actually the same, so after some manipulation we get $$
P(T(X) \le t) = 2 P\left(X \ge \frac 1 {t + 1}, X \in (0, 1/2)\right)
=2 \int_{1 / (t + 1)} ^ {1/2} \ dx = 1 - \frac 2 {t + 1}.
$$
Differentiating to get the pdf gives $$
f(t) = \frac 2 {(t + 1) ^ 2}
$$
for $t \ge 1$ and $f(t) = 0$ otherwise.
As a sanity check, $\int_1 ^ \infty \frac{2t}{(t + 1)^2} \ dt = \infty$ as expected.
