The best way to disprove a statement is to find a counter-example. For first order formulas, the quantifiers are over some universe of discourse $\mathcal{D}$.
For example, $\forall x, P(x)$ says that for every $x$ within the universe of discourse, $x$ has property $P$. Let’s say that our universe of discourse consists of two elements $c_1$ and $c_2$. Then the statement $\forall x, P(x)$ is equivalent to $P(c_1) \wedge P(c_2)$. That is, both elements in the universe of discourse hold property $P$. This equivalence is called the truth-functional expansion.
Let $\mathcal{D} = \{ c_1, \dots, c_n \}$ be the universe of discourse. Then the following equivalences hold: $$ \forall x P(x) \iff P(c_1) \wedge \dots \wedge P(c_n) $$ $$ \exists x P(x) \iff P(c_1) \vee \dots \vee P(c_n) $$
To find a counter-example, you’ll need to find a model in which the formula does not hold. A model consists of a universe of discourse and whether or not each possible predicate holds. For example, let’s take a binary predicate $F$. For a domain of discourse of size two, the following predicates may hold: $F(c_1, c_1), F(c_1, c_2), F(c_2, c_1), F(c_2, c_2)$. For any given model, each of those instantiations can hold or not, amounting to $2^4 = 16$ possible models for a domain of discourse of size $2$ with one binary predicate.
Now let’s find a counter-example for an invalid formula. Consider $\exists x F(x, x) \implies \forall x F(x, x)$. To perform a truth functional expansion, we need to first convert to prenex normal form. This works out to be $\forall x \forall y (F(x, x) \implies F(y, y))$
Details
$$ \begin{align*} \exists x F(x, x) \implies \forall x F(x, x) &\iff \forall x (F(x, x) \implies \forall y F(y, y)) \\ &\iff \forall x \forall y (F(x, x) \implies F(y, y)) \end{align*} $$Let $\mathcal{D} = \{ c_1 \}$. The truth functional expansion is $F(c_1, c_1) \implies F(c_1, c_1)$. This is a tuatlogy so the formula is valid for domain of size 1.
Now let $\mathcal{D} = \{ c_1, c_2 \}$. After removing the tautologies, the truth functional expansion works out to be $(F(c_1, c_1) \implies F(c_2, c_2)) \wedge (F(c_2, c_2) \implies F(c_1, c_1))$
Details
$$ \forall x \forall y (F(x, x) \implies F(y, y)) \\ (\forall y (F(c_1, c_1) \implies F(y, y))) \wedge (\forall y (F(c_2, c_2) \implies F(y, y))) \\ ((F(c_1, c_1) \implies F(c_1, c_1)) \wedge (F(c_1, c_1) \implies F(c_2, c_2))) \wedge ((F(c_2, c_2) \implies F(c_1, c_1)) \wedge (F(c_2, c_2) \implies F(c_2, c_2))) \\ (F(c_1, c_1) \implies F(c_2, c_2)) \wedge (F(c_2, c_2) \implies F(c_1, c_1)) $$
Consider the model $\{ F(c_1, c_1), \neg F(c_1, c_2), \neg F(c_2, c_1), \neg F(c_2, c_2) \}$. This causes the left hand side of the conjunction to fail. Hence the formula is invalid and we have found a counterexample.