Verifying the given answer does not work 
$y=f(x)$, where $f$ satisfies the relation $$f(x+y)=2f(x)+xf(y)+y\sqrt{f(x)}\text{ }\forall x,y\in R$$
  If $f'(0)=0$, then $f(6)=?$ 

The answer given to this question in my textbook (link) is $f(x)=x^2/4$. However, I tried verifying it and it does not seem to be correct. Here's my attempt:
$$\text{LHS}=f(x+y)=(x+y)^2/4=x^2/4+y^2/4+xy/2$$
$$\text{RHS}=2f(x)+xf(y)+y\sqrt{f(x)}\\=2\cdot(x^2/4)+xy^2/4+y\sqrt{x^2/4}\\=x^2/2+xy^2/4+xy/2$$
If we equate $\text{LHS}=\text{RHS}$, which obviously should be true, then $x^2/4+y^2/4=x^2/2+xy^2/4+xy/2$, which is clearly not true.

What mistake am I making in verifying the answer given in my textbook?

EDIT: I have not done the substitution $y=f(x)$, and I can't understand what some commenters below are trying to direct me to.
 A: What they did : they took derivative of the functional equation w.r.t. $x$, put $y=0$, and then solve the differential equation thus generated in $x$.
$$f'(x)=\sqrt {f(x)} \implies \int \frac{f'(x)}{\sqrt{f(x)}} {\rm d} x=\int {\rm d} x \implies f(x)=\frac{x^2}4$$
What they missed is, they can't just bring $\sqrt{f(x)}$ down to denominator on LHS without considering the case $f(x)=0$. So they missed the original, correct solution, right there in that step.
So how come we're getting $f(x) = x^2/4$ at all? 
Well, because this isn't solution of the original functional equation, but it solution of the obtained differential equation in $x$, which must have missed some expression containing $y$ only. Since while original functional equation was differentiated, it was treated as constant and it was treated just like it was never there!
This is just like the way we solve trignometric equations. If we square somewhere, we may possibly generate extra root. If we take square root, we may miss a root. Here, forming DEs is equivalent to squaring. If we get exactly what we need, no problem! But if we get an extra root, we need not to be surprised. 
A: For $y=0$ we obtain $$f(x)=-xf(0),$$ but $x=y=0$ gives $f(0)=0$, which says $f(x)=0$ and $f(6)=0.$
