A differentiable function $f:R\to R$ satisfies the functional equation $f\left(x\right).f\left(y\right)+f(x+y)=e^{x}f\left(y\right)+e^{y}f\left(x\right)+xy\forall x,y\in R.$ If $f\text{'}\left(0\right)=0$ and $f\left(0\right)=0$ then which of the following statements is/are correct?

$\begin{array}{l}f\left(x\right).f\left(y\right)+f(x+y)=e^{x}f\left(y\right)+e^{y}f\left(x\right)+xy\\ \therefore f\left(x\right).f\text{'}\left(y\right)+f\text{'}(x+y)=e^{x}f\text{'}\left(y\right)+e^{y}f\left(x\right)+x\\ Puty=0\\ f\left(x\right).f\text{'}\left(0\right)+f\text{'}\left(x\right)=e^{x}f\text{'}\left(0\right)+f\left(x\right)+x\\ \Rightarrow f\text{'}\left(x\right)-f\left(x\right)=x\Rightarrow f\left(x\right)=e^x-x-1\\ \therefore \left(A\right)\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^2}=\underset{x\to 0}{\lim }\frac{e^x-x-1}{x^2}=\frac{1}{2}\\ \left(B\right)\underset{x}{\overset{x^2}{\int }}tdt={\left(\frac{t2}{2}\right)}_x^{x^2}=\frac{x^4-x^2}{2}\le 0\forall \left|x\right|\le 1\end{array}$

$\begin{array}{l}\left(C\right)F\left(x\right)=f\text{'}\left(x\right)-f\left(x\right)=(e^x-1)-(e^x-x-1)=xwhichisincrea\sin gfunctionso\\ \\ F\left(x_2\right)>F\left(x_1\right)if{x}_2>x_1\\ \left(D\right)Nowf\text{'}\left(x\right)=e^x-1andf\text{'}\text{'}\left(x\right)=e^x>0\forall x\in R\\ \therefore f\text{'}\left(x\right)is\text{\hspace{0.17em}\hspace{0.17em}increasing\hspace{0.17em}\hspace{0.17em}}functionorR\\ \because f\text{'}(-1)=-veandf\text{'}\left(1\right)=+ve\\ \because f\text{'}\left(c\right)=0hasexactlyarerootin(-1,1).\\ i.e.onehorizontal\tan gent.\ln (-1,1)\end{array}$