If $f$ is solution so is $f - cst$ so we may assume for now $f(0) = 0$ Then $f(x^2) = xf(x)$ and equation is equivalent to $f(y)/y = f(x)/x = a$ for $x,y \neq 0$ Solutions are $f(x) = a*x + b$

why can we assume that $ f(0) = 0$ ? I don't understand you...

By "cst" he meant "constant". In other words, a particular function $ F(x)$ is a solution iff. every function of the form $ F(x) + k \; \; \forall k \in \mathbb{R}$ is a solution. Then, for every such family of functions, it suffices to proceed using the member of the family (or the value of $ k$) that makes $ f(0) = 0$.

repeated sol

$f(x^2) - f(y^2) = (x + y)(f(x) - f(y)) (1), \forall x, y \in \mathbb{R}$ Let $y = 0,$ we have $f(x^2) = x(f(x) - f(0)), \forall x \in \mathbb{R}$ From this, $(1)$ will become $x(f(x) - f(0)) - y(f(y) - f(0)) = (x + y)(f(x) - f(y))$ Hence $yf(0) - xf(0) = yf(x) - xf(y)$ In here, let $y$ be an arbitrary nonzero real number $c,$ we have $cf(0) - xf(0) = cf(x) - xf(c)$ Then $f(x) = \dfrac{1}{c}(f(c) - f(0))x + f(0)$ or $f(x) = ax + b, \forall x \in \mathbb{R}, a, b \in \mathbb{R}$ Retry, we see that $f(x) = ax + b$ satisfies $(1)$ In conclusion, we have $f(x) = ax + b, \forall x \in \mathbb{R}, a, b \in \mathbb{R}$