I claim $f(x)=x$ for all $x$ is the only solution. Denote given assertion by $P(x,y)$. $P(0,0)$ yields $f(0)=0$ and $P(x,0)$ implies $f(x^{2019}) = x(f(x))^{2018}$. In particular, $f(x)\ge 0$ for $x\ge 0$ and $f(x)\le 0$ for $x\le 0$. Next, $P(x,-x)$ implies $f(x)^{2018}=f(-x)^{2018}$. Comparing this with $P(-x,0)$, we immediately obtain $f(-x)=-f(x)$ for all $x$. Hence, it suffices to consider $[0,\infty)$. Using $f(x^{2019}) = x(f(x))^{2018}$, we obtain \[ f(x^{2019}+y^{2019}) = f(x^{2019})+ f(y^{2019}). \]Since $a\mapsto a^{2019}$ is surjective, it follows that for all $a,b$; $f(a+b)=f(a)+f(b)$. Namely, $f$ enjoys Cauchy equation. Moreover, it is easily seen that for any $\epsilon>0$, $P(x^{1/2019},\epsilon^{1/2019})$ yields $f(x+\epsilon)\ge f(x)$, that is $f$ is monotonic. Hence, $f(x)=cx$ for some $c$ (all monotonic solutions of Cauchy equation are linear ones). Inserting, and solving for $c$, we get $c=1$, the end.