[จูกัดเหลียง]

Unsure Solution
3.find $f$ with$x,y\in\mathbb{R}$ such that $f(xf(x)+f(y))=f(x)^2+y$
put $y=y-f(x)^2$ get $f(xf(x)+f(y-f(x)))=y$
so $f$ is surjective func. Thus there exist $t$ such that $f(t)=0$
Let $x=t$ we have $f(f(x))=x$
then replace $x$ with $f(x)$ in given function we get $$f(xf(x)+f(y))=f(f(x)f(f(x))+f(y))=f(f(x))^2+y=x^2+y$$
So $x^2+y=f(x)^2+y\rightarrow f(x)=\pm x$