I've been trying to solve this circuit problem(and understand it frankly), and I wanted to double check my thought process with the community helpfully.
After running the circuit out, I have $A+ \bar A B$ for the top half, with the bottom half as $\bar A C$, the issue I have is the last gate is an exclusive-or, so combined(my attempt) was the result $A + \bar A B+ \bar A C$... is this right?
Given this, how do I begin to convert that to a truth table with the exclusive. If I start with for ex, A= 1, B=0, C=0, I end up with 1 as the output?
Following that example, I came up with the outputs as all 1's except for the values A=0, B=0, C=0...
Your suggestion of $$A + \bar A B+ \bar A C$$ would be right if the final gate were a regular inclusive or ($+$), but it isn't; it is an exclusive or ($\oplus$). These are different. I've written the truth tables for both of them below:
$$\begin{array}{cc|cc} A & B & A+B & A\oplus B \\ \hline 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & \color{maroon}{0} \\ \end{array}$$
Because you were treating that final gate as $+$ instead of as $\oplus$, you didn't get the right truth table.
Note that $A+\bar AB$ can be simplified. Do you know the Karnaugh map technique? It is particularly simple in this case.