probability distributions
The continuous random variable $X$ has pdf
$f(x) = 2,$ for $0 < x < 1/2;\;$ $0$ otherwise.
Let $X_1, X_2, X_3$ be independent random variables, all with the same distribution as $X.$ Now, suppose we define $Y = \max\{X_1, X_2, X_3\}.$
Why is $F(y) = Pr(X_1 ≤ y) \cdot Pr(X_2 ≤ y) \cdot Pr(X_3 ≤ y)
= (2y) \cdot (2y) \cdot (2y)$?
The following may get you started, by finding the density functions of $Y$ and $Z.$
For $Z = \max(X_1, X_2, X_3),$ we have the CDF $$F_Z(t) = P(Z \le t) = P(X_1 \le t)P(X_2 \le t)P(X_3 \le t) = t^3,$$ for $0 <t <1.$ [If the maximum is less than $t,$ then so must each of the three independent $X_i$'s be less than $t.$] So $f_Z(t) = 3t^2,$ which is the density function of the distribution $\mathsf{Beta}(3,1).$
Similarly, for the minimum, $F_Y(t) = (1-t)^3,$ for $0 < t < 1,$ and $Y \sim \mathsf{Beta}(1,3).$
For illustration, the plot below shows histograms of simulations of 100,000 realizations of $Y$ and of $Z,$ along with the appropriate Beta densities.
There is no need to work so hard for this question.
Think of it this way. There are only six possible outcomes (except for those outcomes for which at least two of the three random variables are exactly equal, but these occur with probability zero):
$$X_1 < X_2 < X_3 \\ X_1 < X_3 < X_2 \\ X_2 < X_1 < X_3 \\ X_2 < X_3 < X_1 \\ X_3 < X_1 < X_2 \\ X_3 < X_2 < X_1$$ Because each $X_i$ is independent and identically distributed, the probability of any one of these is equal to the other. The underlying distribution makes no difference. Therefore, the answer is $1/6$. Had these not been IID then you'd need to do more work and the answer would depend on how they are distributed, but in this case, the variables are exchangeable.
Best Answer
Comment: Maximum of three uniform random variables.
You begin with independent $X_i \sim Unif(0, 1/2),$ for $i = 1, 2, 3,$ and $Y = \max(X_1, X_2, X_3).$
@AndreNicolas has told you how to obtain $F_Y(y) = P(Y \le y) = (2y)^3,\,$ for $0 < y \le 1/2.$
And you have differentiated to get $f_Y(y) = 24y^2,\,$ for $0 <y \le 1/2.$
Below is R code for a simulation of 100,000 performances of this experiment. That is 100,000 observations of the random variable $Y$. I have made a histogram of them and superimposed your density curve on the histogram. Perhaps you know how to use your density function to find $E(Y)$ and $SD(Y)$ to see how close my approximations are to the exact values (within 2 or 3 places, I'd suppose).