Only using natural deduction prove:
$$\frac{\forall x P(x) \\ \forall x \lnot Q(x) \lor \forall yQ(y) \\ \exists x [P(x) \rightarrow \lnot Q(x)]}{\therefore \forall x \lnot Q}$$
My solution:
$1. \space \forall x P(x) \qquad \qquad \qquad premise\\ 2. \space \forall x \lnot Q(x) \lor \forall y Q(y) \qquad premise\\ 3. \space \exists x [P(x) \rightarrow \lnot Q(x)] \qquad premise \\ \boxed{4. \space x_o \qquad \qquad \qquad \qquad arbitrary \\ \boxed{5. \space y_o \space P(y_o) \rightarrow Q(y_o) \qquad assumption \\ 6. \space \lnot Q(x_o) \ \lor \forall yQ(y) \qquad \forall-elim(2,4) \\ 7. \space \lnot Q(x_o) \lor Q(y_o) \qquad \forall-elim(2,5) \\ \boxed{8.\space y_o\space Q(y_o) \qquad\qquad assumption \\ 9.\space \lnot Q(y_o) \qquad \qquad \rightarrow-elim(5)\\ 10. \space F \qquad \qquad \lnot-elim(8-9)\\ 11. \space \lnot Q(x_o) \qquad \qquad F-elim (10)}\\ \boxed{12. \space \lnot P(x_o) \qquad \qquad assumption \\ 13. \space P(x_o) \qquad \qquad \forall-elim(1,4) \\ 14. \space F \qquad \qquad \lnot-elim(12-13) \\ 15.\space \lnot Q(x_o) \qquad \qquad F-elim(15)}\\ 16. \space \lnot Q(x_o) \qquad \qquad \lor-elim (8-11,12-15)}\\17. \space \lnot Q(x_o) \qquad \qquad \exists-elim(3,5-16)} \\ 18. \forall x\lnot Q(x_o) \qquad \qquad \forall-intro(4-17)$
From the last version I changed the domain on the variables to be seen easier.
Question 1: I think I can change the domain to the second premise because they are two different domains, right?
Question 2: Lines 12 - 15, is this valid?
Question 3: For lines 4 - 7, I said $x_o$ is arbitrary so it can cancel out the for all statement. Is it valid to do it like I did for 6 and 7?
That line is incorrect. In order to apply $\forall E$, Universal Quantifier needs to be the main logical connective.
To use $\lor E$, a statement whose main logical connective is a disjunction needs to appear in previous lines; I cannot see such statement there.
A possible proof could be: $ \def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \def\Ae#1{\qquad\mathbf{\forall\,Elim}\colon #1 \\} \def\Ai#1{\qquad\mathbf{\forall\,Intro}\colon #1 \\} \def\Ee#1{\qquad\mathbf{\exists\,Elim}\colon #1 \\} \def\Ei#1{\qquad\mathbf{\exists I}\colon #1 \\} \def\R#1{\qquad\mathbf{R}\colon #1 \\} \def\ci#1{\qquad\mathbf{\land\,Intro}\colon #1 \\} \def\ce#1{\qquad\mathbf{\land\,Elim}\colon #1 \\} \def\oe#1{\qquad\mathbf{\lor\,Elim}\colon #1 \\} \def\ii#1{\qquad\mathbf{\to Intro}\colon #1 \\} \def\ie#1{\qquad\mathbf{\to Elim}\colon #1 \\} \def\be#1{\qquad\mathbf{\leftrightarrow E}\colon #1 \\} \def\bi#1{\qquad\mathbf{\leftrightarrow I}\colon #1 \\} \def\fi#1{\qquad\mathbf{\bot\,Intro}\colon #1 \\} \def\qi#1{\qquad\mathbf{=I}\\} \def\qe#1{\qquad\mathbf{=E}\colon #1 \\} \def\ne#1{\qquad\mathbf{\neg E}\colon #1 \\} \def\ni#1{\qquad\mathbf{\neg I}\colon #1 \\} \def\IP#1{\qquad\mathbf{IP}\colon #1 \\} \def\X#1{\qquad\mathbf{\bot\,Elim}\colon #1 \\} \def\DNE#1{\qquad\mathbf{DNE}\colon #1 \\} $
$ \fitch{1.\,\forall xP(x)\\ 2.\,\forall x\lnot Q(x) \lor \forall xQ(x)\\ 3.\,\exists x(P(x) \to \lnot Q(x)) }{ \fitch{4.\,[a]}{ \fitch{5.\,\forall x \lnot Q(x)}{ 6.\,\lnot Q(a) \Ae{6} }\\ \fitch{7.\,\forall x Q(x)}{ \fitch{8.\,[b]\,P(b) \to ¬Q(b)}{ 9.\,P(b) \Ae{1} 10.\,\lnot Q(b) \ie{8,9} 11.\,Q(b) \Ae{7} 12.\,\bot \fi{10,11} }\\ 13.\,\bot \Ee{3,8-12} 14.\,\lnot Q(a) \X{13} }\\ 15.\,\lnot Q(a) \oe{2,5-6,7-14} }\\ 16.\,\forall x~\lnot Q(x) \Ai{4,15} } $