So, you know that the equation is going to look like
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$
You already figured out that $a=-b$. So now you need to figure out what values of $a$ and $b$ will make the foci be at the given point $(4,0)$ and $(-4,0)$.
Since you know that $a=-b$, then you know that $a^2=b^2$, so you can rewrite the equation as
$$\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$$
or, by clearing denominators (multiplying through by $a^2$)
$$x^2 - y^2 = a^2.$$
Now: where does the hyperbola intersect the $x$ axis? If you have a point $(X,0)$ in the hyperbola, then plugging it in you will have
$$X^2 = a^2.$$
So if you can determine where the hyperbola intersects the $x$-axis, you can use that to figure out the value of $a$, and so the value of $b$, and so the equation for the hyperbola.
Note: There are many, many, many ways to finish the problem, depending on what you know about hyperbolas in general (which is precisely why I asked you in comments what you knew).
Added. One possible way to finish the problem is to use the eccentricity of the hyperbola. The distance from the center to the foci is $a\varepsilon$, where $\varepsilon$ is the eccentricity of the hyperbola. The eccentricity is always equal to
$$\varepsilon = \sqrt{ 1 + \frac{b^2}{a^2}}$$
so here, since $a=b$, that means that the eccentricity is $\sqrt{2}$. Therefore, since the distance from the center to the foci is $4$, this tells you that $4 = a\varepsilon = a\sqrt{2}$. You now know what $a$ is, and thus, you know what $a^2=b^2$ is, and so you know the equation of the hyperbola.
Added 2. Another possibility you discuss in the comments. You have $a$ is the distance from the center to the vertices; $b$ is the distance from the center to the conjugate axis. These distances satisfy $c^2 = a^2+b^2$, where $c$ is the value $4$ that you have. Since $a^2=b^2$, then you have $c^2=2a^2$, or $c = \sqrt{2a^2} = \sqrt{2}|a| = a\sqrt{2}$. You already know what $c$ is, so now you can solve for $a$. Once you know $a$, you also know $b$ (since $a=-b$) which gives you the information you need.
Your calculation is OK, and shows that taking $a>c$ leads to $2c=d.$ But in this case, every point on the $x$ axis to the right of $c$ (or to the left of $-c$) will satisfy the locus definition, since the calculation will be (e.g. for $x>c$) $$|x+c|-|x-c|=(x+c)-(x-c)=2c=d.$$ We certainly don't want to call something a hyperbola which includes such rays of the $x$ axis.
Best Answer
As $y=3$ contains the foci, it also contains the major axis
If the equation is $\dfrac{(x-\alpha)^2}{a^2}-\dfrac{(y-\beta)^2}{b^2}=1$
As the center is the midpoint of the foci, we have $2\alpha=-1+3,2\beta=3+3$
Now the coordinates of the foci are $(\alpha\pm a\varepsilon,\beta)$
So, $1+2a=3,1-2a=-1\implies a=1$
We know $b^2=a^2(\varepsilon^2-1)$
Hope you can take it form here