Finding probability density function of a conditional expectation I´m solving the next exercise of a probability book about conditional expectation:
Let $(X,Y)$ a discrete random vector with probability density function given by
$$f_{x,y}(x,y)=(x+y)/36\quad  x,y\in\{1,2,3\}$$ and zero in other case.
a) Find the probability density function of the random variable $E(X|Y).$
b) Check the formula $E(E(X|Y))=E(X)=78/36.$
I've begun computing marginal density for variable $Y.$ My computations lead me to get $$f_{Y}(y)=\dfrac{2+y}{12}\quad\text{if}\quad y\in\{1,2,3\}$$ and zero in other case.
Then I use the definition of $E(X|Y=y),$ so doing the computations I get $$E(X|Y=y)=\dfrac{6y+14}{6+3y}.$$ 
Due to the above, we conclude that $E(X|Y)=\dfrac{6Y+14}{6+3Y}.$ Then we want to calculate $$P(\dfrac{6Y+14}{6+3Y}=y)$$ which is equivalent to compute $f_{Y}(\frac{6y-14}{6-3y}),$ but here is my problem. I don't know how to calculate that term, because random variable $Y$ is discrete, so the point of evaluation must be a natural number where marginal density of $Y$ takes these values.
How can I compute that value? I would like that $\frac{6y-14}{6-3y}=z$ for every value of $z\in\{1,2,3\},$ but when I do that, the new values of $y$ are rationals.
What am I doing wrong? Is there another way to solve this easily?
Any kind of help is thanked in advanced. 

$\require{cancel}$You've the right calculations for $$f_{Y}(y)=\frac{2+y}{12} \text{ if } y\in\{1,2,3\}$$ and $$E(X \mid Y=y)=\frac{6y+14}{6+3y} \text{ if } y\in\{1,2,3\}.$$
The interpretation of $E(E(X \mid Y))$ should be
\begin{align}
& E(E(X \mid Y)) \\
=& \sum_{y = 1}^3 E(X \mid Y = y) P(Y = y) \\
=& \sum_{y = 1}^3 \frac{6y + 14}{\cancelto{3}{6 + 3y}} \frac{\cancel{2+y}}{12} \\
=& \sum_{y = 1}^3 \frac{6y + 14}{36} \\
=& \frac{6 \times 6 + 3 \times 14}{36} \\
=& \frac{78}{36}.
\end{align}
\begin{align}
& E(X) \\
=& \sum_{x = 1}^3 \sum_{y = 1}^3 x f_{X,Y}(x,y) \\
=& \sum_{x = 1}^3 \sum_{y = 1}^3 x \frac{x+y}{36} \\
=& \sum_{x = 1}^3 \frac{3x^2+6x}{36} \\
=& \sum_{x = 1}^3 \frac{x^2+2x}{12} \\
=& \frac{1+4+9+2\times6}{12} \\
=& \frac{26}{12} \\
=& \frac{78}{36}.
\end{align}
This is not a coincidence.  In fact,
\begin{align}
& E(E(X \mid Y)) \\
=& \sum_y E(X \mid Y = y) P(Y = y) \\
=& \sum_y \sum_x x P(X \mid Y = y) P(Y = y) \\
=& \sum_y \sum_x x \frac{f_{X,Y}(x,y)}{\cancel{f_Y(y)}} \cancel{f_Y(y)} \\
=& \sum_y \sum_x x f_{X,Y}(x,y) \\
=& E(X).
\end{align}

Edit in response to OP's comment
\begin{align}
f_Y(y) =& \sum_{x = 1}^3 f_{X,Y}(x,y) \\
=& \sum_{x = 1}^3 \frac{x+y}{36} \\
=& \frac{1+2+3+3y}{36} \\
=& \frac{2+y}{12}
\end{align}
