Solving functional equation $f(x)\cdot f(y)-xy=f(x)+f(y)-1$ As in the title. Substituting $y=x$ we get:
$$ f(x)^2-x^2=2f(x)-1 $$
after rearranging, we get:
$$ f(x)(f(x)-2)=(x+1)(x-1) $$
And I rather cannot assume that e.g. $f(x)=x+1$ and $f(x)-2=x-1$, so what should I do now?

Note that in your first step you had the following equation:
$$f(x)^2-x^2=2f(x)-1$$
Instead of factoring, if we rearrange all the terms on one side we get:
$$f(x)^2 - 2f(x) - x^2 + 1 = 0$$
Note we have a quadratic, with $f(x)$ acting like our $x$ and the term $1-x^2$ being our constant term.
Solving, we get:
$$f(x) = \frac{2 \pm \sqrt{4 -4(1-x^2)}}{2}$$
$$ = \frac{2 \pm \sqrt{4 - 4 + 4x^2)}}{2}$$
$$ = \frac{2 \pm \sqrt{4x^2}}{2}$$
$$ = \frac{2 \pm 2x}{2}$$
Therefore, we get $2$ solutions for $f(x)$:
$$f(x) = \frac{2 + 2x}{2} = 1+x$$
$$f(x) = \frac{2 - 2x}{2} = 1-x$$
