$x=1:f(y^2)=y^2f(1) \to f(x)=xf(1)$ for $x \geq 0$ $x=-1: f(-y^2)=y^2f(-1) \to f(x)=-xf(-1)$ for $x<0$ $f(2008)=f(-2008) \to f(1)=f(-1)=c$ So $f(x)=|x|c$

Setting $y=1$ we get $f(x^2)=x^2f(1)$ and setting $y=-1$ we get $f(-x^2)=x^2f(-1)$. Using condition, $f(1)=f(-1)$ and so $x\mapsto c|x|$, which works.