$$
\lnot (\forall x \lnot Px \lor \forall x \lnot Qx)) \Rightarrow \lnot (\forall x (\lnot Px \lor \lnot Qx ))\\$$
is not correct.
It is equivalent to
$$
\varphi ~ =: ~\forall x \lnot Px \lor \forall x \lnot Qx \Leftarrow \forall x (\lnot Px \lor \lnot Qx ) ~ := \psi\\$$
Consider your examplary $\{P,Q\}$-structure $A$ over the universe $\{0,1\}$ with $P=\{0\}$ and $Q=\{1\}$.
Now $A \models \psi$ but $A \not \models \varphi$.
This the schema
$\forall x A x \lor \forall x B x \Rightarrow \forall x (A x \lor B x)$ is correct but you have been using it the otherway around which is generally not a correct implication.
Yes, you can prove it with "usual" proof systems : Resolution, Tableaux or with Natural Deduction :
1) $∀z∀x∃y(R(z,y)∨¬P(x))$ --- premise
2) $∃xP(x)$ --- assumed [a]
3) $P(w)$ --- assumed [b] for $\exists$-elimination
4) $∃y(R(z,y)∨¬P(w))$ --- from 1) by $\forall$-elimination twice
5) $R(z,y)∨¬P(w)$ --- assumed [c] for $\exists$-elimination
6) $P(w) \to R(z,y)$ --- from 5) by tautological equivalence : $(p \to q) \leftrightarrow (\lnot p \lor q)$
7) $R(z,y)$ --- from 3) and 6) by $\to$-elimination
8) $\exists yR(z,y)$ --- from 7) by $\exists$-introduction
9) $\forall x \exists yR(x,y)$ --- from 8) by $\forall$-introduction : $x$ is not free in any "open" assumptions
$y$ is not free in 9); thus, we can apply $\exists$-elimination with 4), 5) and 9) and conclude with :
10) $\forall x \exists yR(x,y)$, discharging assumption [c].
In the same way, from 2), 3) and 10), discharging assumption [b] and concluding with :
11) $\forall x \exists yR(x,y)$.
Finally :
$∃xP(x) \to \forall x \exists yR(x,y)$ --- from 2) and 11) by $\to$-introduction, discharging assumption [a].
With a final application of $\to$-introduction we conclude with :
$∀z∀x∃y(R(z,y)∨¬P(x)) \to (∃xP(x) \to ∀x∃yR(x,y))$.
Best Answer
$\def\boxit#1{\bbox[lemonchiffon,0.5ex]{#1}}$We have premises of $\boxit{\forall x~\forall y:(P(x,y)\to Q(x))}$ and $\boxit{\forall x~\exists y:P(x,y)}$. Should we take an arbitrary variable, $\boxit a$, then we infer from the second premise that there is a witness variable, call it $\boxit b$, which satisfies $\boxit{P(a,b)}$. For these variables, we also infer from the first premise that $\boxit{P(a,b)\to Q(a)}$ will be satisfied. Thus, by modus ponens, we infer that $\boxit{Q(a)}$ is satisfied. Since $\boxit b$ does not occur in this statement, and $\boxit a$ is arbitrary, we have therefore shown that $\boxit{\forall x:Q(x)}$ is entailed by these premises.
$$\def\fitch#1#2{~~~~{\begin{array}{|l}#1\\\hline#2\end{array}}}\fitch{~~1.~\forall x\,\forall y:(P(x,y)\to Q(x))\hspace{3.5ex}\textsf{Premise}\\~~2.~\forall x\,\exists y:P(x,y)\hspace{14ex}\textsf{Premise}}{\fitch{~~3.~\boxed a\hspace{23.5ex}\textsf{Assumption (Arbitrary)}}{~~4.~\forall y:(P(a,y)\to Q(a))\hspace{4ex}\textsf{Universal Elimination, 1}\\~~5.~\exists y:P(a,y)\hspace{14.5ex}\textsf{Universal Elimination, 2}\\\fitch{~~6.~\boxed b~P(a,b)\hspace{13.5ex}\textsf{Assumption (Witness)}}{~~7.~P(a,b)\to Q(a)\hspace{8ex}\textsf{Universal Elimination, 4}\\~~8.~Q(a)\hspace{18.5ex}\textsf{Conditional Elimination, 6, 7}}\\~~9.~Q(a)\hspace{21.5ex}\textsf{Existential Elimination 5, 6-8}}\\10.~\forall x:Q(x)\hspace{19.75ex}\textsf{Universal Introduction, 3-9}}$$