f(x) is differentiable function sattisfying the relationship $f^2\left(x\right)+f^2\left(y\right)+2(xy-1)=f^2(x+y)\mathrm{\forall }x,y\in R$ Also $f\left(x\right)>0\mathrm{\forall }x\in R$ and $f\left(\sqrt{2}\right)=2.$ Then which of the following statements is/are correct about f(x)?

$\text{ Differentiable w.r.t. ' }x\text{ ' }$

$2f\left(x\right)f^{\mathrm{\prime }}\left(x\right)+2y=2f(x+y)f^{\mathrm{\prime }}(x+y)$

$$\begin{gathered} put x=0 \\ k+y=f^{\mathrm{\prime }}\left(y\right)f\left(y\right) \end{gathered}$$

$$\begin{gathered} integrating on both sides \\ ky+\frac{y^2}{2}=\frac{f^2\left(y\right)}{2}+c \end{gathered}$$

$If x=0,y=0 then f^2\left(0\right)=2\Rightarrow f\left(0\right)=\sqrt{2}$

$If y=0 then c=\frac{f^2\left(0\right)}{2}\Rightarrow c=1$

$put y=\sqrt{2 }then$

$k\sqrt{2}+1=\frac{4}{2}-1$

$k\sqrt{2}=0\Rightarrow k=0$

$\frac{y^2}{2}=\frac{f^2\left(y\right)}{2}-1$

$f^2\left(y\right)=y^2+2$

$f\left(y\right)=\sqrt{y^2+2}$

$f\left(x\right)=\sqrt{x^2+2}$

Given relations are satisfied by the function $f\left(x\right)=\sqrt{x^2+2}$