Let $f:\mathbb{R}→\mathbb{R}$ such that: $$f(xy)=xf(y)+yf(x), f(x+y)=f(x^{2021})+f(y^{2021}), \forall x, y\in \mathbb{R}$$ Calculate $f(\sqrt{2020})$.
So far I found out that $f(x)$ is an additive function that has the form of $f(x)=ax\log_c(x)$, with $a, c$ is constant, $a\neq 0$, $c>0$. I figured it out from the first equation. Now I don't know what to do with the second equation, neither as how to calculate.
For $x\neq 0$ set $g(x) := \frac {f(x)}x$. Then the first equation shows that $g(xy) = g(x)+g(y)$, whenever $x,y\neq 0$. In particular, $g(2x) = g(2) + g(x)$ and $g(x^n) = ng(x)$ for $n\in\mathbb N$. The second equation shows that \begin{align} g(2) + g(x) &= g(2x) = \frac{f(2x)}{2x} = \frac{2f(x^{2021})}{2x} = \frac{f(x^{2021})}{x^{2021}}\cdot x^{2020} = x^{2020}g(x^{2021})\\ &= 2021x^{2020}g(x). \end{align} Plugging in $x=2021^{-(2020)^{-1}}$ in this equation gives $g(2) = 0$ and therefore $f(x)=0$ for $x\neq 0$. The first equation also yields $f(0)=0$. Therefore, $f\equiv 0$ is the only solution.