As you didn't show your results, I'll redo the computation.
Let us substitue the RHS of the two middle equations in the fourth, squared:
$$d^2(x^2+b)y^2=(y^2+c)x^2,$$ or
$$(d^2b-x^2)y^2=(c-d^2)x^2.$$
Then, from the first equation,
$$y^2=(a-x)^2$$
and finally, eliminating $y$,
$$(c-d^2)x^2=(d^2b-x^2)(a-x)^2.$$
This confirms that the problem is of the quartic degree and has an analytical solution, but the formulas are terrible and hardly manageable by hand.
And as the polynomial has three independent coefficients, there is no reason that any simplification occurs and you won't find any simpler analytical resolution (otherwise, you would revolutionize the Galoi's theory).
The other option is to resort to numerical methods, but whether this is feasible by hand is unsure, and it is only doable for particular values of the parameters $a,b,c,d$.
I directly solved the three equations. From first equation $$ z = \frac{1-x}{xy}-1 = \frac{1}{xy}-\frac{1}{y}-1$$, substituting this in the second equation, $$ y + y\left(\frac{1-x}{xy}-1 \right) + (1-x-xy)=2 \implies xy = \frac{1}{x}-x-2. $$
This implies $xyz = 1-x-xy = 3-\frac{1}{x}$ and $z= \frac{1-x}{\frac{1}{x}-x-2}-1 = \frac{3x-1}{1-x^2-2x}$. Substituting this in the third equation results in a cubic equation, $$\left(\frac{3x-1}{1-x^2-2x}\right)+\left(\frac{3x^2-x}{1-x^2-2x}\right)+\left(3-\frac{1}{x}\right)=4 \\ \implies \frac{3x^2+2x-1}{1-x^2-2x}-\frac{1}{x}=1 \implies 3x^2+2x-1-\frac{1}{x}+x+2 = 1-x^2-2x\\ \implies \boxed{4x^3+5x^2-1=0}.
$$
This has three solutions $x = -1,-\frac{1}{8}-\frac{\sqrt{17}}{8},-\frac{1}{8}+\frac{\sqrt{17}}{8}$. The maximum of $xyz = 3-\frac{1}{x}$ is attained when $x=-\frac{1}{8}-\frac{\sqrt{17}}{8}$ and it is $$ \boxed{\frac{1}{2}(5+\sqrt{17})} $$
Best Answer
Approach 1: Adding up the equations, $30 (a + b + c + d) = a^2 + b^2 + c ^2 + d^2 \geq \frac{ (a+b+c+d) ^2 } { 4} $.
Since $ a+b + c + d > 0$, we can divide by it and conclude that hence $ 120 \geq a+b+c+d$.
WLOG, let $ a= \min (a, b, c, d)$.
Then $ a^2 = 9b + 10c + 11d \geq 9a + 10 a + 11a = 30a$, so $ a \geq 30$.
The only possible tuple which satisfies both conditions is $(30,30,30,30)$, which we can easily verify is a solution.
Approach 2: (Fill in the gaps yourself) Similar to the 2nd step from above, show that
$$ 30 \leq \min (a, b, c, d) \leq \max (a, b, c, d ) \leq 30. $$
Note: