It can be shown that if

$$\begin{aligned} \forall x .(\exists y. p(x,y) \implies q(x)) \end{aligned}$$

holds, then

$$\begin{aligned} \forall x .\forall y.(p(x,y) \implies q(x)) \end{aligned}$$

holds. Do you see why ?

Turns out it's due to the simple reason that

$$\begin{aligned} \forall x .(\exists y. p(x,y) \implies q(x)) \end{aligned}$$

is not the same as

$$\begin{aligned} \forall x .(\exists y. (p(x,y) \implies q(x))). \end{aligned}$$

In particular, the sentence

$$\begin{aligned} \exists y. p(x,y) \implies q(x) \end{aligned}$$

does not mean there exists $y$ that the implication holds. Rather, it means if there exists $y$ so that $p(x,y)$ holds, then $q(x)$ holds. In other words, the quantifier "exists y" applies only to the antecedent and not to the consequence.