Proof of implication for inference I'm gonna prove if $x$ is not free in $\psi$, then $(\phi \rightarrow \psi)\models (\exists x \phi \rightarrow \psi) $.
Here is my argument:
According to definition of rule of inference of QR, if $x$ is free, then one can deduce $(\exists x \phi \rightarrow \psi)$ from $\phi \rightarrow \psi$. Mathematically: $(\phi \rightarrow \psi)\vdash (\exists x \phi \rightarrow \psi) $. Furthermore, from soundness theorem, we know that, if $a \vdash b$ then $a \models b$. So, according to truth of $(\phi \rightarrow \psi)\vdash (\exists x \phi \rightarrow \psi) $, $(\phi \rightarrow \psi)\models (\exists x \phi \rightarrow \psi) $ holds.
Is this sort of argument valid?
If one says (you must prove that "if $x$ is free, then one can deduce $(\exists x \phi \rightarrow \psi)$ from $\phi \rightarrow \psi$", I don't know the argument for that.)
UPDATE: 
$(\phi \rightarrow \psi) \models (\exists x \phi \rightarrow \psi)$
$(\phi \rightarrow \psi) \models \neg(\exists x \phi) \vee \psi$
$(\phi \rightarrow \psi) \models (\forall x (\neg \phi)) \vee \psi$
$(\neg\phi \vee \psi) \models (\forall x (\neg \phi)) \vee \psi$
$(\neg\phi \vee \psi) \models (\forall x (\neg \phi)) \vee \psi$
By universal instantiation:
$(\neg\phi \vee \psi) \models (\neg \phi) \vee \psi$
Which is true.

 A: So two things:
First, you are trying to prove that this logical entailment holds on the basis of an existing inference rule plus soundness. OK, that should work ... If you know that this specific inference rule is sound.  Do you know that?  I mean, yes, there are many sound systems of logic (indeed, presumably most systems of logic are sound) .. But just because we are presented with some system and some inference rule doesn't mean it is sound; we would have to prove that it is sound, which means we would need to prove exactly the entailment you want to prove. So unless someone told you that this specific inference rule is sound, this would become a circular argument. Indeed, I suspect that the whole goal of the exercise you are presented with is to prove the very soundness of this very inference rule, meaning that your approach would indeed be circular (you'd assume the proof system that includes this inference is sound in order to prove that this inference rule is sound). So, you most likely need to use formal semantics to prove this.
Second, about the actual soundness of this inference rule (or the validity of the entailment) ... I must say it looks very weird: it says that $x$ is not free in $\psi$ .. But the inference rule really doesn't do anything with $\psi$; instead, it quantifies the $\phi$. Meaning that if $x$ is a free variable in $\phi$, then we have a big problem since that would mean that we have a different set of variables free between the left hand side and the right hand side... in a way, we would be comparing apples with oranges! So, I think a mistake was made here, and that they should have said that $x$ is not a free variable in $\phi$ .. And then the entailment actually holds!
EDIT
OK, so I had this last part wrong: with a proper semantics, you can make sense of (and prove!) $\phi \rightarrow \psi \vDash \exists x \phi \rightarrow \psi$, even if $\phi$ contains free variables (see Mauro's answer). 
A: Hint
See Ch.2.5 Soundness : our rules of inference will preserve truth, i.e. for a rule $⟨Γ,θ⟩ : Γ⊨θ$.
And see Th.2.5.2 for the proof that the rules are sound.
The Exercise asks to prove that the second QR rule :

$\langle \{ \phi \to \psi \} , (\exists x) \phi) \to \psi \rangle$, $x$ not free in $\psi$,

preserves truth.
Assume a structure $\mathfrak A$ such that $\mathfrak A \vDash (\phi \to \psi )$, i.e. for any assignment function $s$ : $\mathfrak A \vDash (\phi \to \psi )[s]$.
We have to prove that : $\mathfrak A \vDash ((\exists x)\phi \to \psi )[s']$, for every $s'$.
Case (i) : If $\mathfrak A \vDash \psi[s']$, it's done.
Thus, consider :
Case (ii) : $\mathfrak A \nvDash \psi[s']$.
We want that $\mathfrak A \nvDash (\exists x)\phi[s']$, i.e. that $\mathfrak A \nvDash \phi[s'[x|a]]$ for some $a \in \text {dom}(\mathfrak A)$.
But we know that : $\mathfrak A \vDash (\phi \to \psi )[s]$ for every $s$, and thus also : $\mathfrak A \vDash (\phi \to \psi )[s'[x|a]]$.
But $x$ is not free in $\psi$ and thus $s'$ and $s'[x|a]$ agree on the free variables of $\psi$; thus, from $\mathfrak A \nvDash \psi[s']$ we have also : $\mathfrak A \nvDash \psi[s'[x|a]]$, for some $a$.
This, due to  $\mathfrak A \vDash (\phi \to \psi )[s'[x|a]]$ above, implies that : $\mathfrak A \nvDash \phi[s'[x|a]]$, for some $a$, i.e.:

$\mathfrak A \nvDash (\exists x)\phi[s']$.

