$f_E(e) = f_{E\mid G,A}(e\mid g,a)$ since they are both $1/3$. I also noted that the probability of any $g,a$ in the respective sample space is $1/6$ which covers the full sample space since g x a = 6 (Hopefully I did this right). I am having issues with this and think that I may be mixing values up or approaching this in the wrong way.
Would there be another way of going about this (without obviously manually writing the whole sample space) in order to confirm that this holds true? stat inferential question
Suppose $f_{\text{Education | Gender, Age}}(e\mid g,a) = 1/3$ for each $e$ for any $g\in\{0,1\}$ and $a\in\{20,30,40\}$, and that $f_{\text{Gender,Age}}(g,a)=1/6$ for any such $g, a$. Are Education and (Gender, Age) independent?
Nope, that is the best way to go.
Since for all six combinations of gender and age, the probability mass functions (for any $e$) are constant, thence we can simply calculate:
$$\begin{align}f_{\text{Education}}(e) &= \sum_{\langle g,a\rangle\in\{0,1\}{\times}\{20,30,40\}}f_{\text{Education|Gender,Age}}(e\mid g,a)\,f_{\text{Gender,Age}}(g,a)\\&=6\cdot(1/3)\cdot(1/6)\\&=1/3\\&=f_{\text{Education|Gender,Age}}(e\mid g,a)\end{align}$$
As this holds for any $e$ and all $g,a$ (as provided), this establishes the independence of Education from (Gender, Age).