We are given a function $f$ and need to determine its form under certain conditions. The initial setup suggests introducing $a = -b$ in order to analyze the functional equation $f(b^2) = f(b) + f(-b) - f(b)f(-b)$. The goal is to ensure that $f$ is non-decreasing.

According to the problem statement, we want $f(b^2) \geq f(b)$, leading to $f(-b)[1 - f(b)] \geq 0$. This condition can be satisfied in two main cases. We analyze these cases separately to derive the behavior of the function $f$.

### Case 1: $f(b) = 1$ and $f(-b) = 0$

If $f(b) = 1$, then for $a < 0$, $f(a) = 0$. Substituting into the functional equation gives $1 = f(a+b-ab)$. For any $a, b$ such that $a+b-ab = x$ with $x \geq 1$, there is always a solution $b \geq 1$ for any fixed $a$ and $x$. Hence, if $f(b) = 1$, it implies:
$$a \leq 0 \implies f(a) = 0,$$
$$0 < a < 1 \implies f \text{ is non-decreasing and } 0 < f(a) < 1,$$
$$b \geq 1 \implies f(b) = 1.$$

### Case 2: $f(-b) = 0$

In this case, for $a < 0$, we again have $f(a) = 0$ and $b \geq 1$ implies $f(b) = C \geq 1$. For $0 < a < 1$, from $f(a) + C - f(a)C = C$, we find $f(a)(1 - C) = 0$. Therefore, if $C = 1$, the function is as in Case 1. If $C > 1$, $f(a) = 0$. So, the function satisfies:
$$a < 1 \implies f(a) = 0,$$
$$f(1) = 1,$$
$$b > 1 \implies f(b) = C.$$

### General Form of the Function

From the analysis above, we deduce the structure of the function $f$ must adhere to different behaviors based on intervals of $x$. Thus, we express $f$ as:
$$f(x) = \begin{cases} -(1-x)^c + 1 & \text{if } x < 1 \\ 1 & \text{if } x = 1 \\ k(x-1)^c + 1 & \text{if } x > 1 \end{cases},$$
where $c \geq 0$ and $k > 0$ are constants.

Thus, the function is determined by the conditions on $x$ and respective parameters $c$ and $k$.

Final answer:
$$\boxed{f(x) = \begin{cases} -(1-x)^c + 1 & \text{if } x < 1 \\ 1 & \text{if } x = 1 \\ k(x-1)^c + 1 & \text{if } x > 1 \end{cases}, \text{ for some real constants } c \geq 0, k > 0}$$