Injective or surjective?

Not injective. $f(-1)=f(1)=1$ although $-1\ne1$.

Not surjective. $f(x)=x^2\ge0$ for every $x$, so $-1$ has no preimage.

A bijection. Let $g:[0,+\infty)\to[0,+\infty)$ be the restriction $g(x)=x^2$. For $y\ge0$, the equation $x^2=y$ has exactly one solution in $[0,+\infty)$, namely $x=\sqrt y$. So $g$ is bijective, with inverse $y\mapsto\sqrt y$.