I don't know why, but these questions are always easier on continuous random variables...anyway, can someone please look at my solution below and see if it is correct/incorrect, especially with all the domains and such?
The p.d.f. of discrete $X$ is given by $$f_{X}(x) = \begin{cases} x/6, & x=1,2,3 \\ 0, & \text{elsewhere} \end{cases}$$
The definition of the c.d.f. of $X$ is $F_{X}(x) = \underset{x \leq w}{\sum}f_{X}(w)$. This our c.d.f. is given by $$F_{X}(x) =\begin{cases} 0, & x<1 \\ 1/6, & 1 \leq x<2\\ 3/6, & 2\leq x <3 \\ 1, & 3 \leq x \end{cases}$$
We define $Y = X^{2}$. This $Y$ has its own space $\mathscr B = \{1, 4, 9\}$. Thus the c.d.f. of $Y$ will be found by $$F_{Y}(y) = Pr(Y \leq y) = Pr(X^{2} \leq y) = Pr(-\sqrt{y} \leq X \leq \sqrt{y})$$
But the $Pr(-\sqrt{y} \leq X \leq \sqrt{y}) = F_{X}(\sqrt{y}) - F_{X}(-\sqrt{y})$ where $F_{X}(-\sqrt{y}) = 0$, and so $F_{Y}(y) = F_{X}(\sqrt{y})$ which is given by $$F_{Y}(y) = \begin{cases} 0, & y<1 \\ 1/6, & 1 \leq y<4\\ 3/6, & 4\leq y <9 \\ 1, & 9 \leq y \end{cases}$$
Finally, the p.d.f. of $Y$ is given by $$f_{Y}(y) = \begin{cases} {\sqrt y\over 6}, & y=1,4,9 \\ 0, & \text{elsewhere} \end{cases}$$.
Let me know guys, thank you!
When handling change of variables for probability mass functions (pmf), you do not need to use cumulative distributions. There is no scaling of density to account for; the discrete points in the support have probability mass.
$$\begin{align}f_X(x) &= \begin{cases}x/6 &:& x\in\{1,3,6\}\\[1ex]0 &:& \text{else}\end{cases}\\[3ex]f_{Y}(y) &= f_X(\surd y)&&\text{since }Y=X^2\\[1ex]&=\begin{cases}(\surd y)/6&:& \surd y\in\{1,2,3\}\\[1ex]0&:&\text{else}\end{cases}\\[1ex]&=\begin{cases}(\surd y)/6&:& y\in\{1,4,9\}\\[1ex]0&:&\text{else}\end{cases}\end{align}$$