The problem states that real numbers $a$, $b$, and $c$ satisfy the conditions:

$abc = -1, \quad a + b + c = 4$

It asks to find $a^2 + b^2 + c^2$ based on the following equation:

$\frac{a}{a^2 - 3a - 1} + \frac{b}{b^2 - 3b - 1} + \frac{c}{c^2 - 3c - 1} = \frac{4}{9}$

Let's begin by analyzing the given information and solving for $a^2 + b^2 + c^2$. I'll work through the details of simplifying the equation and incorporating the conditions provided.We have the expanded form of $(a + b + c)^2$:

$16 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

Next, we need to express $ab + bc + ca$. By using the symmetric identities derived from the sum and product of roots in cubic equations, we'll substitute these into the equation to isolate $a^2 + b^2 + c^2$. Let me continue from here.The expression for $ab + bc + ca$ is:

$ab + bc + ca = -\frac{1}{2}(a^2 + b^2 + c^2) + 8$

Now, we can substitute this result back into the equation $16 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ and solve for $a^2 + b^2 + c^2$. Let me do that next.