OK, the cumulative distribution function (CDF) is
$$
P(X \leq x) = F(x)
= \begin{cases}
0 & \text{for } x < 0, \\
9x^2 & \text{for }0 \le x \le 1/3, \\
1 & \text{for }x > 1/3.
\end{cases}
$$
and the probability density function is
$$
f(x) = F'(x)
= \begin{cases}
0 & \text{for }x < 0, \\
18x & \text{for }0 < x < 1/3, \\
0 & \text{for }x > 1/3.
\end{cases}
$$
This is continuous in the sense that the CDF is continuous, and also in the stronger sense that 100% of the probability comes from integrating a density function (absolute continuity with respect to Lebesgue measure). The mode is 1/3. The mean is less than 1/3.
The right "tail" runs from 1/3 to 1 only because you've declared (0,1) the interval of interest by fiat. Or if you like, take $(1 - 10^{100})$ times this density and add $10^{100}$ times the uniform density on the interval $(1/3,1)$. Then the support of the distribution is $[0,1]$ so it's not just by fiat.