1. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$, the following holds:

$$
x f(x+y)+y f(y-x)=f\left(x^{2}+y^{2}\right)
$$
Solution. Let's note that all functions of the form $f(x)=c x$, where $c$ is a real constant, satisfy the equation.

With $P(x, y)$ we denote the statement

$$
x f(x+y)+y f(y-x)=f\left(x^{2}+y^{2}\right)
$$

From $P(0,0)$ we get $f(0)=0$ and together with $P(x, 0)$, we conclude that $x f(x)=$ $f\left(x^{2}\right)$ for all $x \in \mathbb{R}$. From $P(-x, 0)$ we also have that $-x f(-x)=f\left(x^{2}\right)=x f(x)$ for all $x \in \mathbb{R}$. Therefore, if $x \neq 0$ then $f(-x)=-f(x)$. Considering that $f(0)=0$, it follows that $f(-x)=-f(x)$ for all $x \in \mathbb{R}$, i.e., $f$ is an odd function.

From $P(y, x)$ we get:

$$
y f(y+x)+x f(x-y)=f\left(y^{2}+x^{2}\right)=f\left(x^{2}+y^{2}\right)=x f(x+y)+y f(y-x)
$$

According to the oddness of $f$, $f(x-y)=-f(y-x)$ holds, which means that:

$$
y f(x+y)-x f(y-x)=x f(x+y)+y f(y-x)
$$

Let's fix $t \in \mathbb{R}$ and substitute $x=\frac{t-1}{2}$ and $y=\frac{t+1}{2}$ into the last equation. We get

$$
f(t)=t \cdot f(1), \quad \text { for every } t \in \mathbb{R}
$$

Therefore, all solutions are given by $f(x)=c x$, where $c$ is a real constant.