It seems that you start with ∀x P(x) and instantiate x with y. However, ∀x P(x) is not given: it is an assumption of an assumption and thus has to be proved. It is also not clear why you need to instantiate x since you can immediately derive ∃x Q(x) by MP and from there derive the conclusion.is arbitray, so
If, on the other hand, you assume P(y), then you can conclude ∀x P(x) because the assumption contains y free. Also, it is not clear where you close the assumption P(y).
To say something more precise, you have to say what formalism you are using, even informally (e.g., natural deduction or sequent calculus).
This conclusion can't be proved without the law of excluded middle. Suppose that ∀x P(x) is true. Then ∃x Q(x) from the assumption, and so ∃x (P(x) -> Q(x)). If ∀x P(x) is false, then there exists an x0 such that P(x0) is false. Therefore, P(x0) -> Q(x0) is true.