 Solution

True. The map $(s,t)\mapsto (s^2,t^2)$ is a homeomorphism of $[0,1{]}^2$. Hence every continuous $f(s,t)$ can be written as $g(s^2,t^2)$ for continuous
$g(u,v)=f(\sqrt{u},\sqrt{v})$. Bernstein polynomials approximate $g(u,v)$ uniformly, so polynomials in $s^2,t^2$ approximate $f$ uniformly. The assumed even moments annihilate all such polynomials; approximating $f$ and passing to $\int f^2$ proves $f=0$.