If f : R → R be a differentiable function, such that $f(x+2 y) = f(x)+f(2 y)+4 x y$ for all x, y ∈ R then,

f'(0) = f'(1) - 2

We have,

$f(x+2 y)=f(x)+f(2 y)+4 x y$  for all x, y ∈ R

Putting x = y = 0, we get f(0) = 0

Now,

$f(x+2 y)=f(x)+f(2 y)+4 x y$

$\Rightarrow \frac{f(x+2 y)-f(x)}{2 y}=2 x+\frac{f(2 y)}{2 y}$

$\Rightarrow \lim\limits_{y \rightarrow 0} \frac{f(x+2 y)-f(x)}{2 y}=\lim\limits_{y \rightarrow 0}\left\{2 x+\frac{f(2 y)-f(0)}{2 y}\right\}$

⇒ f'(x) = 2x + f'(0) for all x

$\Rightarrow f'(1)=2+f'(0)$             [Replacing x by 1]

$\Rightarrow f'(0)=f'(1)-2$