Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ 
If $x,y\in \mathbb{R}$ and $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ and $f'(0)=0\;,$ Then $f(x)$ is

$\bf{My\; Try::}$ Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)}{h}$$
Now Put $x=y=0$ in  $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}\;,$$ We get $f(0)=0$
So we get $f(0)=0$
So $$f'(x) = \sqrt{f(x)}+\lim_{h\rightarrow 0}\frac{f(h)}{h}=\sqrt{f(x)}$$
So $$\int\frac{f'(x)}{\sqrt{f(x)}}dx = 1\int dx\Rightarrow 2\sqrt{f(x)}=x+c$$
Now Put $x=\;,$ We get $c=0$
So we get $2\sqrt{f(x)}=x\Rightarrow 4f(x)=x^2\Rightarrow \displaystyle f(x)=\frac{x^2}{4}$
Can we solve it some short way, If yes then please explain here, Thanks
 A: $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}=f(y+x)=f(y)+f(x)+x \sqrt{f(y)}$$
Subtracting $f(x)+f(y)$ from each side and squaring , we have that $$y^2f(x)=x^2f(y) \Leftrightarrow \frac{f(x)}{x^2}=\frac{f(y)}{y^2}$$
So we have $\frac{f(x)}{x^2}$ is a constant function. 
Now put $f(x)=cx^2$ in the original equation, where $c$ is a constant. Simplifying gives us that $$2cx=\sqrt{c} |x|$$
Note that if $c \neq 0$, $c$ will take different values depending on the value of $x$, This is a contradiction, as $c$ is a constant. So we have $c=0$. Thus, $f(x)=0$ is the only solution. In order for $\frac{x^2}{4}=f(x)$, to be a solution we must have a constraint that $x \ge 0$, or the functional equation should be altered so: $$f(x+y)=f(x)+f(y)+y\operatorname{sgn}(x)\sqrt{f(x)}$$
A: The above answers didn't see, that, by squaring, an incorrect solution is added. Indeed, there is no nontrivial solution:
In order for the functional equation to be well defined for each $x,y\in\mathbb R$, we need $f\geq 0$. Let's assume that there is $z\in\mathbb R$ with $f(z)>0$.
Because $f(0) = f(0)+f(0)$, we have $f(0)=0$, hence $z\neq 0$.
Then, for $x=z,~y=-z$, we obtain
$$f(0) = f(z)+f(-z)-z\sqrt{f(z)}.$$
On the other hand, for $x=-z,~y=z$, we obtain
$$f(0)=f(-z)+f(z)+z\sqrt{f(-z)}.$$
Subtracting those equations, we get
$$0 = z\sqrt{f(-z)}+z\sqrt{f(z)}.$$
Since $z\neq 0$, this yields $0=\sqrt{f(-z)}+\sqrt{f(z)}\geq \sqrt{f(z)}>0$, contradiction.
An other possibility is, to see that the only nontrivial solution could be $x^2/4$ as shown above and then check that it is indeed not a solution to the equation.
A: I like writing
$$\frac{f(x+y)-f(x)}y=\frac{f(y)}y+\sqrt{f(x)}$$
Then take the limit as $y\rightarrow0$ to get
$$f^{\prime}(x)=f^{\prime}(0)+\sqrt{f(x)}$$
(assuming differentiability and noting that $f(0)=0$ so we can apply L'Hopital's rule) so that if $f^{\prime}(0)=0$, then
$$\frac{df}{\sqrt f}=dx$$
$$2\sqrt{f(x)}=x+C$$
From the original equation it can be seen that $f(0)=0$ so $C=0$ and it follows that
$$f(x)=\frac{x^2}4$$
Notice that this technique allows you to get implicit solutions even if $f^{\prime}(0)\ne0$.  
EDIT: But such solutions are invalid because if $f^{\prime}(0)\ne0$, then since $f(0)=0$, there would be some points near $x=0$ where $f(x)<0$ which would be wrong because then $\sqrt{f(x)}\notin\mathbb{R}$.
