Then the value of $ [f(2)] $ where [.] represents the greatest integer function is? 
A differentiable function f is satisfying the relation $$f(x+y) = f(x) + f(y) + 2xy(x+y) - \dfrac{1}{3} $$ $ \forall $  $  x , y $ belongs to $\Re$ and $$lim_{h \to 0} \dfrac{3f(h)-1}{6h} = \dfrac{2}{3}. $$Then the value of $ [f(2)] $ where [.] represents the greatest integer function is?

I think  $g(x+y)=g(x)+g(y)$ then we use the limit condition, I need help, and If someone can complete my solution I'll appreciated 
 A: Setting $$f(x)=g(x)+\frac{2x^3+1}3$$ we get $$g(x+y)=g(x)+g(y)$$ and so, since continuous, $g(x)=ax$
Using then the limit condition, we get $a=\frac 43$ and so $$f(x)=\frac{2x^3+4x+1}3$$ and $$f(2)=\frac{25}3$$
And $$\boxed{\left\lfloor(f(2)\right\rfloor=8}$$
A: Using the formula $$\displaystyle f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}...............(1)$$
Now Given $$\displaystyle f(x+y) = f(x)+f(y)+2xy(x+y)-\frac{1}{3}$$
So Put $x=h\;$ We get $$\displaystyle f(x+h) = f(x)+f(h)+2xh(x+h)-\frac{1}{3}$$
So $$\displaystyle f'(x) = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+2xh(x+h)-\frac{1}{3}-f(x)}{h}$$
$$\displaystyle f'(x) = \lim_{h\rightarrow 0}2x(x+h)+\lim_{h\rightarrow 0}\frac{3f(h)-1}{3h}  = 2x^2+\frac{4}{3}$$
So $$\displaystyle \int f'(x)dx  = \int 2 x^2+\frac{4}{3}\int 1dx$$
So $$\displaystyle f(x) = \frac{2}{3}x^3+\frac{4}{3}x+\mathcal{C}$$
Now Put $x=y=0$ in original functional equation, we get $$\displaystyle f(0) =\frac{1}{3}$$
So Put $x=0$ in $$\displaystyle f(x) = \frac{2}{3}x^3+\frac{4}{3}x+\mathcal{C}$$
We get value of $\mathcal{C} = \frac{1}{3}$
So we get $$\displaystyle f(x) = \frac{2}{3}x^3+\frac{4}{3}x+\frac{1}{3}$$
Put $x=2\;,$ We get $\displaystyle f(2) = \frac{25}{3}$
So we get  $\lfloor f(x)\rfloor = 8$
