[ThE-dArK-lOrD]

Note that this solution is not mine
Let $P(x,y)$ be the assertion $f(x^2)+f(y)=f(x^2+y+xf(4y))$
Case $1)$ If $f(u)=0$ for some $u>0$
$P(\sqrt u,u)$ give us $f(2u+\sqrt uf(4u))=0$, note that $2u+\sqrt{u}f(4u)\geq 2u$,
Let then $x\ge 0$ and $u>x$ such that $f(u)=0$. Quadratic $X^2+Xf(4x)+x-u$ has one real positive root $v$
$P(v,x)$ give us $f(v^2)+f(x)=f(v^2+x+vf(4x))=f(u)=0$ and so $f(v^2)=f(x)=0$
And so $f(x)=0$ for all $x\in \mathbb{R}^+_0$
Case $2)$ $f(x)>0$ $\forall x>0$
Suppose there exist $a,b$ such that $f(a)=f(b)$ and $a<b$
Quadratic $x^2+xf(4a)+a-b$ has one real positive root $v$
$P(v,a)$ give us $f(v^2)+f(a)=f(v^2+a+vf(4a))=f(b)=f(a)$ and so $f(v^2)=0$, impossible.
So $f(x)$ is injective.
$P(x,1),P(1,x^2)$ give us $f(x^2+1+xf(4))=f(1+x^2+f(4x^2))$ give us $f(4x^2)=xf(4)$
And so $f(x)=a\sqrt x$ , then we easily get that $f(x)=\sqrt{x}$ for all $x\in \mathbb{R}^+_0$