Let $f$ be a differential function satisfying the relation $f(x+y)=f(x)+f(y) - 2xy + (e^x -1)(e^y -1)$$ \ \forall x , y \in\mathbb R $ and $f'(0)=1$
My work
Putting $y=0$ $$f(x)=f(x) + f(0)$$
$$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ $$f'(x)=\lim_{h\to 0} \frac{f(x)+f(h) - 2xh + (e^x -1)(e^h -1)-f(x)}{h}$$ $$f'(x)=\lim_{h\to 0} \frac{f(h) - 2xh + (e^x -1)(e^h -1)}{h}$$ How to predict things after that?
As $f$ is differentiable, $$ \lim_{h \to 0} \frac{f(h)}{h}=f'(0)=1 $$ Also, $$\lim_{h \to 0} \frac{-2xh+(e^x-1)(e^h-1)}{h}=-2x+(e^x-1)\lim_{h \to 0} \frac{e^h-1}{h}=-2x+e^x-1$$
Putting this together, $$ f'(x)=1-2x+e^x-1=-2x+e^x $$ and you can integrate to find $f$.
Alternatively, write the functional equation as $$f(x+y)-e^{x+y}+(x+y)^2+1=f(x)-e^x+x^2+1 + f(y) - e^y +y^2+1$$
Substitute $g(x)=f(x)-e^x+x^2+1$. This gives
$$g(x+y)=g(x)+g(y)$$
which is the Cauchy functional equation. Because $f$ is differentiable at zero, all solutions are given by $g(x)=cx$ for some $c \in \mathbb R$.
So $f(x)=e^x+cx-x^2-1$. So $f'(x) = e^x+c-2x$, so $f'(0)=1+c=1$, hence $c=0$.
So $f(x)=e^x-x^2-1$.
