Answer
$(0, 0)$
Work Step by Step
The vertex-form of a quadratic function is $y=a(x-h)^{2}+k$ where the vertex is the opposite of $h$ and $k$, or $(h, k)$. Given the function from the problem, $f(x)=x^{2}$, written in vertex form this function looks like $f(x)=1(x-0)^{2}+0$. Therefore the vertex is at $(0, 0)$.