It looks pretty symmetric

Let's focus on the first equation and get an expression for $a$ in terms of $b$ :

$\begin{aligned} \dfrac{a}{b} &= 3 + 3ab - b^2 \\ a &= 3b + 3ab^2 - b^3 \\ a(1 - 3b^2) &= 3b - b^3 \\ a &= \dfrac{3b - b^3}{1 - 3b^2}. \end{aligned}$

Seeing this expression reminds us of the tangent triple angle formula: $\tan 3 \theta = \dfrac{3 \tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}.$ Indeed, since $a, b, c, d$ are all real numbers, they can all by expressed as a tangent of some angle, so we think to use a tangent substitution to make solving the system easier.

Let $a = \tan A.$ Then,

$\dfrac{3a - a^3}{1 - 3a^2} = \dfrac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} = \tan 3A.$

Thus, $d = \tan 3A.$ By continually substituting, we get $c = \tan 9A,$ $b = \tan 27A,$ and finally, $a = \tan 81A = \tan A.$ If the tangent of two angles is the same, then they differ by an integer multiple of $\pi,$ so we have $81A = A + k\pi$ for some integer $k,$ or $A = \dfrac{k\pi}{80}.$

For every possible value of $A,$ we get a unique solution to the system, since all the rest of the angles are determined from $A.$ Since the tangent function has a period of $\pi,$ we only need to consider $-40 < k < 40.$ The only value that does not work is $k = 0,$ since that gives us $A = B = C = D = 0$ and $a = b = c = d = 0,$ which is clearly not allowed. Any other value of $k$ in that interval is permissible, giving us $(80 - 1) - 1 = \boxed{78}$ unique solutions to the system of equations.