> 
> $y=f(x)$, 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 $f(6)=?$ 
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The answer given to this question in my textbook ([link](https://www.cengage.co.in/category/test-prep/jee-advanced/mathematics/course/mathematics-for-joint-entrance-examination-jee-advanced-calculus-y1)) 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?
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**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.