to find the distribution function Let $X_1,X_2,X_3$  be mutually  stochastically independent random variables and let each of them  have  the following density function:

$f(x)= 2x$ when $0\leq x\leq 1$ and $f(x)=0$ elsewhere.

Let Y be the random variable defined as, $Y=\max(X_1, X_2, X_3)$, find (a) the  distribution function and (b) the  probability  function of the random variable $Y$.
I saw somewhere that the distribution function of such variable to be given by  $F_Y(y)=P(Y\leq y)=P(\max (X_1,X_2,X_3)\leq y)$. 
If  this  is true, how  may I apply it in this question to find (a) and (b) above?
 A: You were almost finished. The cumulative distribution function $F(y)$ of $Y$ is then the probability that all the $X_i$ are $\le y$.  For any $i$, and $0\le y\le 1$,
$$P(X_i \le y)=\int_0^y 2x\,dx=y^2.$$
So, for $0\le y \le 1$, the probability that all the $X_i$ are $\le y$ is, by independence, $(y^2)^3$.  
We conclude that $F(y)=0$ if $y<0$, $F(y)=y^6$ if $0\le y\le 1$, and $F(y)=1$ if $y>1$.  For the density function of $Y$, differentiate.
Comment: Let $W$ be the minimum of the $X_i$. Then $P(W \le w)$ is $1$ minus the probability that all the $X_i$ are $\ge w$.  For any $i$, and $0 \le w\le 1$, $P(X_i>w)=1-w^2$.  It follows that 
$$P(W \le w)=1-(1-w^2)^3.$$
A: The answer follows just from what you have told us:
Let $F_Y$ be the distribution function of $Y$ and let $F$ be the distribution function of $X_1$ ( and hence that of $X_2, X_3$ as these are identically distributed )
$$F_Y(y) = P(Y \leq y) = P(\max(X_1, X_2, X_3) \leq y)$$
Since the maximum of three numbers is less than $y$, each of them must also be less than $y$. And, since the $X_1, X_2, X_3$ are independent random variables,
$$F_Y(y) = P(X_1 \leq y, X_2 \leq y, X_3 \leq y)$$
$$F_Y(y) = P(X_1 \leq y)\cdot P(X_2 \leq y)\cdot P(X_2 \leq y)$$
$$F_Y(y)=  (F(y))^3$$
I am sure you'll figure out what $F$ is, for yourself.
