Functional Bashing!

Putting $x=y$ ,

$f(2x)=2f(x)-[f(x)]^2$

Since it has been given that $\displaystyle f'\left(0\right)=-1$ , it means that the function is atleast once differentiable. Therefore, differentiating both sides, we have:

$\large \displaystyle 2f'(2x)=2f'(x)-2f(x)f'(x)$

$\large \displaystyle \implies 2f'(0)=2f'(0)-2f(0)f'(0) \implies 2(-1)=2(-1)-2f(0)(-1) \implies \boxed{f(0) = 0}$

Now, obtain the derivative of $f$ using first principle:

$\large \displaystyle f'(x) = \lim_{h \to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h} = \lim_{h \to 0} \frac{f\left(x\right)+f\left(h\right)-f\left(x\right)f\left(h\right)-f\left(x\right)}{h} = (1 - f(x)) \lim_{h \to 0} \frac{f\left(h\right)}{h} = (1 - f(x)) \lim_{h \to 0} \frac{f\left(h\right)-f\left(0\right)}{h-0} = (1 - f(x)) f'(0) = f(x) - 1$

$\large \displaystyle \implies f'(x) = f(x) - 1$

Solving this differential equation,

$\large \displaystyle \implies \frac{df\left(x\right)}{f\left(x\right)-1}=dx \implies \ln\left(f\left(x\right)-1\right)=x+C \implies f\left(x\right)=C'e^x+1=Ce^x+1$

Since $\large \displaystyle f(0) = 0$ , $\large \displaystyle C=-1$ . Therefore, $\large \displaystyle \boxed{f(x) = 1 - e^x}$ .

$\large \displaystyle \large \displaystyle \int_0^1 1 - e^x dx = \boxed{2-e}$