Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$

$$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}=f(y+x)=f(y)+f(x)+x \sqrt{f(y)}$$ 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: $$f(x+y)=f(x)+f(y)+y\operatorname{sgn}(x)\sqrt{f(x)}$$

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.