A differentiable function $\text{⨍ }$ is satisfying the relation $\text{⨍ (x+y)=⨍ (x)+⨍ (y)+2xy(x+y)-}\frac{1}{3}\text{,∀x, y∈R}$ and $\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{3\text{⨍ (h)-1}}{6\mathrm{h}}=\frac{2}{3}.$ Then the value of $\left[\text{⨍ (2)}\right]$ is where $\left[.\right]$ is G.I.F

$$\begin{gathered} {\text{⨍ }}^1\text{(x)=}\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{\text{⨍ (x+h)-⨍ (x)}}{\mathrm{h}} \\ {\text{⨍ }}^1\text{(x)=}\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{\text{⨍ (x)+⨍ (h)+2xh(x+h)-}{\displaystyle \frac{1}{3}}\text{-(⨍ (x)+⨍ (0)-}{\displaystyle \frac{1}{3}}\text{)}}{\mathrm{h}} \\ {\text{⨍ }}^1\text{(x)=}\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{\text{⨍ (h)-⨍ (0)}}{\mathrm{h}}{\text{+2x}}^2{\text{=⨍ }}^1{\text{(0)+2x}}^2 \end{gathered}$$

Given ${\text{⨍ }}^1\text{(x)=}\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{3\text{⨍ (h)-1}}{6\mathrm{h}}\text{=}\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{\text{⨍ (h)-}{\displaystyle \frac{1}{3}}}{2\mathrm{h}}\text{=}\underset{\mathrm{h}\to 0}{\mathrm{Lt}}\frac{\text{⨍ (h)-⨍ (0)}}{2\mathrm{h}}$

$\Rightarrow \frac{{\text{⨍ }}^1\text{(0)}}{2}=\frac{2}{3}\Rightarrow \therefore {\text{⨍ }}^1\text{(0)=}\frac{4}{3}$

$={\text{⨍ }}^1\text{(x)=}\frac{4}{3}{\text{+2x}}^2$

$\text{⨍ (x)=λ+}\frac{4}{3}\text{x+}\frac{2{\mathrm{x}}^3}{3}$ (or) $⨍ \left(0\right)=\mathrm{\lambda }=\frac{1}{3}$

$⨍ \left(\mathrm{x}\right)=\frac{2{\mathrm{x}}^3}{3}+\frac{4}{3}\mathrm{x}+\frac{1}{3}$ (or) $⨍ \left(2\right)=\frac{25}{3}$

$\Rightarrow \left[⨍ \left(2\right)\right]=8$