3. Rewrite the original equation in the form

$$
16 z^{2}+4 x y z+\left(y^{2}-3\right)=0
$$

This is a quadratic equation in terms of z. It has a solution if $D / 4=4 x^{2} y^{2}-16\left(y^{2}-3\right) \geq 0$. After transformations, we obtain the inequality

$$
y^{2}\left(x^{2}-4\right)+12 \geq 0
$$

If $x^{2}-4 \geq 0$, then the first term on the left side of the inequality is non-negative, and consequently, the left side is greater than or equal to 12 for any $y$. If $x^{2}-4<0$, then the first term on the left side of the inequality is negative, and by choosing $y$, we can make the left side of the inequality negative. Therefore, the condition of the problem is satisfied by $x$ for which $x^{2}-4 \geq 0$. From this, we find $x \in(-\infty ;-2] \cup[2 ;+\infty)$.
Answer: $\quad x \in(-\infty ;-2] \cup[2 ;+\infty)$.