An algebra problem by A Former Brilliant Member

Let denotes the functional equation by $P(x,y)$ as follows. We need to find the constant term of $f(x)$ or $f(0)$ .

$\begin{array}{ll} P(x,y): & f(x-f(y)) = f(f(y)) + xf(y) + f(x) - 1 \\ P(0,x): & f(-f(x)) = f(f(x)) + f(0) - 1 \\ P(0,1): & f(-{\color{#3D99F6} f(1)}) = f({\color{#3D99F6} f(1)}) + f(0) - 1 & \small \color{#3D99F6} \text{Note that }f(1) = 0 \\ & f({\color{#3D99F6} 0}) = f({\color{#3D99F6} 0}) + f(0) - 1 \\ \implies & f(0) = \boxed 1 \end{array}$

The question asks for "the constant term", which I'm taking to mean $f(0)$ .
Let $x = 0, y = 1$ .

Since $f(y) = 0$ , $x - f(y) = x$ . Thus $f(0) = f(0-f(1))$ .

By the given relationship: $f(0 - f(1)) = f(f(1)) + 0f(1) + f(0) - 1$ $= f(0) + 0 + f(0) - 1$ .

In summary $f(0) = 2f(0) - 1$ , i.e. $f(0) = 1$ .