Confusing in limits of bivarite distribution and contradiction in related question

The rule of $3y_1$ only holds when $0 \le y_2 \le y_1 \le 1$, it takes zero otherwise.

Let's describe the region, what are the values that $y_2$ can take? It takes value between $0$ and $\frac13$.

What are the values that $y_1$ can take in term of $y_2$, it takes value between $y_2$ and $\frac12$.

The limit of the integral should be

\begin{align} \int_0^\frac13 \int_{y_2}^\frac12 3y_1 \,\, dy_1 dy_2 &= \int_0^\frac13 \frac32y_1^2|_{y_2}^\frac12 \, dy_2 \\ &= \frac32\int_0^\frac13 \frac14 - y_2^2 \, dy_2\\ &= \frac32 \left(\frac1{12} - \frac1{81} \right)\\ &=\frac12 \left(\frac14 -\frac1{27}\right) \end{align}

For continuous variable, the random variable takes a particular value with probability $0$, hence the equality does not matter.