A function $f$, defined for all $x, y \in R$ is such that $f\left(1\right)=2; f\left(2\right)=8$ and $f\left(x+y\right)-kxy=f\left(x\right)+2y^2$, where $k$ is some constant.  If $f\left(x+y\right)f\left(\frac{1}{x+y}\right)=k$ for $x+y\ne 0$ then k =

$f\left(1\right)=2 ; f\left(2\right)=8 \forall x;y\in R$

and $f\left(x+y\right)-kxy=f\left(x\right)+2y^2$

put $x=1; y=1$

$$\begin{gathered} \therefore f\left(2\right)-k=f\left(1\right)+2 \\ \Rightarrow 8-k=2+2 \\ \Rightarrow k=4 \\ \therefore f\left(x+y\right)=4xy+f\left(x\right)+2y^2 \end{gathered}$$

put $y=-x$

$$\begin{gathered} f\left(0\right)=2x^2+f\left(x\right)-4x^2 \\ f\left(0\right)=f\left(x\right)-2x^2 \end{gathered}$$

put $x=1 ; y=-1$

$$\begin{gathered} f\left(0\right)=-4+f\left(1\right)+2 \Rightarrow f\left(0\right)=-4+2+2=0 \\ \therefore f\left(x\right)=2x^2 \\ f\left(x+y\right)=2x^2+2y^2+4xy \\ f\left(x+y\right)=2{\left(x+y\right)}^2 \Rightarrow f\left(t\right)=2t^2 \\ f\left(\frac{1}{x+y}\right)=2·\frac{1}{{\left(x+y\right)}^2} \end{gathered}$$

$\therefore f\left(x+y\right)·f\left(\frac{1}{x+y}\right)=\frac{2{\left(x+y\right)}^2\times 2}{{\left(x+y\right)}^2}=4$ hence proved