If \(f\left(x^{2}+f(y)\right)=x f(x)+y\) for all non-negative integers \(x\) and \(y\), then the value of \([f(0)]^{2}+f(0)\) equals _________.

x=0: f(f(y))=y. Let f(0)=a, so f(a)=0. y=a: f(x^2)=x f(x)+a; x=a gives f(a^2)=a*0+a=a, then f(a)=f(f(a^2))=a^2, so a^2=0, a=0. Hence f(0)^2+f(0)=0 (f(x)=x satisfies the equation).