I have no idea how to get tabular to work on here so the table isn't rendering.
A joint PMF $p_{X, Y}[i, j]$ has the values shown in table. Determine the conditional PMF $p_{Y|X}$. Are the random variables independent? $$ \begin{array}{|c|c|c|c|} \hline & j = 1 & j = 2 & j = 3\\ \hline i = 1 & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{2}{10}\\ \hline i = 2 & \tfrac{1}{20} & \tfrac{1}{20} & \tfrac{1}{10}\\ \hline i = 3 & \tfrac{3}{10} & \tfrac{1}{20} & \tfrac{1}{20}\\ \hline \end{array} $$
When I follow the formula, the conditional probabilities aren't summing to one.
The $p_{Y|X = x_i}[j|x_i] = \frac{p_{X, Y}[x_i, j]}{p_X[x_i]}$ for all $j$ and similarly for $p_{X|Y = y_j}$. \begin{alignat*}{4} p_{Y|X = 1}[j|1] &= \begin{cases} \frac{3}{10}, & j = 1\\ \frac{3}{10}, & j = 2\\ \frac{6}{10}, & j = 3 \end{cases} & \qquad & p_{Y|X = 2}[j|2] &&={} \begin{cases} \frac{3}{20}, & j = 1\\ \frac{3}{20}, & j = 2\\ \frac{3}{10}, & j = 3 \end{cases}\\ p_{Y|X = 3}[j|3] &= \begin{cases} \frac{9}{10}, & j = 1\\ \frac{3}{20}, & j = 2\\ \frac{3}{20}, & j = 3 \end{cases} & \qquad & p_{X|Y = 1}[i|1] &&={} \begin{cases} \frac{3}{10}, & i = 1\\ \frac{3}{20}, & i = 2\\ \frac{9}{10}, & i = 3 \end{cases}\\ p_{X|Y = 2}[i|2] &= \begin{cases} \frac{3}{10}, & i = 1\\ \frac{3}{20}, & i = 2\\ \frac{3}{20}, & i = 3 \end{cases} & \qquad & p_{X|Y = 3}[i|3] &&={} \begin{cases} \frac{6}{10}, & i = 1\\ \frac{3}{10}, & i = 2\\ \frac{3}{20}, & i = 3 \end{cases} \end{alignat*}
I'm not sure where you are getting those figures. The marginal probability of $X$ is the sum of the joint probabilities for all values of $Y$ at particular values of $X$.
$\begin{align} \text{Thus:} \\ \because p_X(i) & = \sum_{j=1}^3 p_{X,Y}(i,j) \\ &= p_{X,Y}(i,1)+p_{X,Y}(i,2)+p_{X,Y}(i,3) \\[1ex] \therefore p_X(1) &= p_{X,Y}(1,1)+p_{X,Y}(1,2)+p_{X,Y}(1,3) \\ &= \tfrac 1{10}+\tfrac 1{10}+\tfrac 2{10} \\ &= \tfrac 2 5 \\ \text{Similarly:} \\[2ex] \because p_{Y\mid X}(j\mid i)&= \frac{p_{X,Y}(i, j)}{p_X(i)} \\[2ex] \therefore p_{Y\mid X}(j\mid 1)&= \frac{p_{X,Y}(1, j)}{p_X(1)} \\ & = \frac{5}{2}\begin{cases}\frac 1{10} & :j=1\\\frac 1{10} &:j=2\\\frac 2{10} &:j=3\end{cases} \\ & = \begin{cases}\frac 1{4} & :j=1\\\frac 1{4} &:j=2\\\frac 1{2} &:j=3\end{cases} \end{align}$
You can do the rest now, right?