Given an $\mathcal{L}-$language, prove if it exists or not a deduction for: $$\exists x_1 \exists x_2 \neg \varphi \vdash \neg \exists x_1 \exists x_2 \varphi $$
My idea is that if it exists a syntactic deduction, by the Soundness Theorem it follows the semantic implication, so a set composed of the first formula and the negative of the second is not satisfiable. I have tried this but I am not getting the fact, possible ideas would be appreciated.
First of all, you should recognize that the statement is not true in general. So, we are trying to disprove it.
Therefore, we want to find a model which satisfies both the affirmative of the left-hand side and the negative of the right-hand side. That's actually fairly easy, since we just want two elements such that $\varphi$ is true for them and two elements such that $\varphi$ is false for them. For example, let $\mathcal{M}$ be a model with $\lvert \mathcal{M} \rvert = \{a, b, c, d\}$ and $\varphi^{\mathcal{M}} = \{(a, b)\}$. Then $a, b$ make $\varphi$ true and $c, d$ make $\varphi$ false. Done.
Here I assumed, though, that $\varphi$ was a relational symbol — you haven't specified what it was. If the problem asks if this is true for all formulas $\varphi$ of $\mathcal{L}$, then the above is enough, as we have found a formula for which it is not true. There do exist, though, some formulas for which the statement is true; for example, take $\varphi = \top$.