$∀x(P(x)→Q(x)) \ and \ ∀xP(x)→∀xQ(x)$ are logically equivalent if and only if ,
$(∀x(P(x)→Q(x)) \ → \ ∀xP(x)→∀xQ(x)) \ and \ ((∀xP(x)→∀xQ(x)) \ → \ ∀x(P(x)→Q(x)))$.
Let's prove the implications in parts.
$(∀x(P(x)→Q(x)) \ → \ ∀xP(x)→∀xQ(x))$
We need to prove that this implication is Valid.
The implication is not valid if we can find a True -> False case.
RHS is false if $∀xP(x)→∀xQ(x)$ is false thus $∀xP(x)$ must be True , and for some $x \ Q(x)$ must be false.
Since $P(x)$ is true for every $x$ of the domain , and $Q(x)$ must be false for some $x$ , thus LHS can never be true since there will be atleast one case of $True->False$ on LHS , thus making the LHS completely false.
Thus we couldn't prove the implication to be invalid , thus the implication is valid.
Now let's prove the other part of the implication.
$((∀xP(x)→∀xQ(x)) \ → \ ∀x(P(x)→Q(x)))$.
Here too we need to show atleast one $True->False$ case.
RHS if false if $∀x(P(x)→Q(x)))$ is false , thus there must exist one such $x$ for which $P(x)$ is true and $Q(x)$ is false.
LHS is true , if $((∀xP(x)→∀xQ(x))$ is true . Let there be some $x$ for which both $P(x) \ and \ Q(x)$ is false , this will make LHS True .
Thus we were able to find a $True->False$ situation , making the implication Invalid.
Thus both are $NOT \ EQUIVALENT$