I claim all such functions are form $f(x)=cx$, where $c\in\mathbb{R}$ is arbitrary. Denote the given assertion by $P(x,y)$. Note that $P(0,y)$ gives $f(y^2)=yf(y)$. In particular $f(0)=0$ and $f((-y)^2)=-yf(-y)$ so that $f(-y)=-f(y)$ for any $y$. Now, \begin{align*} P(x,y)&\Rightarrow xf(x+y) + yf(y-x) = f(x^2+y^2) \\ P(y,x)&\Rightarrow yf(x+y) + xf(x-y) = f(x^2+y^2). \end{align*}As $f(y-x)=-f(x-y)$, we get \[ (x-y)f(x+y) = (x+y)f(x-y) \]for any $x,y$. Now we are ready for conclusion. Take any $r_1,r_2\ne 0$ and let $x,y$ be such that $x+y=r_1$ and $x-y=r_2$. We then get immediately $f(r_1)/r_1=f(r_2)/r_2$ for any $r_1,r_2\ne 0$. So, there is a $c$ such that $f(x)=cx$ for any $x\ne 0$. Clearly this extends to all $x$ as $f(0)=0$, which completes the solution.

a trivial problem just need to change x into -x.

Natsuki_kobayashi_486 wrote: a trivial problem just need to change x into -x. That's why it's the first of the five problems.

steppewolf wrote: Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ we have: $$xf(x+y)+yf(y-x) = f(x^2+y^2)\,.$$Proposed by Nikola Velov $$xf(x+y)+yf(y-x) = f(x^2+y^2)...(\alpha)$$In $(\alpha) x=y=0:$ $$\Rightarrow f(0)=0$$In $(\alpha) x=0:$ $$\Rightarrow yf(y)=f(y^2)$$$$\Rightarrow f \text{ is odd}$$In $(\alpha) x=y, y=x:$ $$yf(x+y)+xf(x-y)=f(x^2+y^2)...(\beta)$$By $(\alpha)$ and $(\beta):$ $$\Rightarrow yf(x+y)+xf(x-y)=xf(x+y)+yf(y-x)$$$$\Rightarrow (x+y)f(x-y)=(x-y)f(x+y)$$$x+y=a, x-y=b:$ $$\Rightarrow af(b)=bf(a), \forall a,b \in \mathbb{R}...(\theta)$$In $(\theta) a=1:$ $$\Rightarrow f(b)=bf(1), \forall b\in\mathbb{R}$$Replacing in $(\alpha):$ $$f(x)=xc, c\text{ is constant and }x\in\mathbb{R} \text{ is the only solution}$$