I assume that the equation you've given is of interest to you, so here is my interpretation of Sasha's comment. The equation under consideration has degree $2$ in all three variables, so a compactification of this surface in $\mathbb CP^1\times \mathbb CP^1\times \mathbb CP^1$ is a divisor of degree $(2,2,2)$. So if by any chance it is smooth, then indeed it is a K3 surface. It might happen, however that the surface has singularities in $\mathbb C^3$ or in one of $7=2^3-1$ other charts. So one needs to find the solution of $F_x=F_y=F_z=F=0$ in all these charts. Probably this can be done with a help of some program. Once these singularities are found one needs to check if these are Du Val or not (for this they need to be isolated of course). This should not be super hard since the degree in each variable is $\le 2$.
PS. As Jianrong is pointing out, the surface is singular at the point $(0,0,-3)$. In order to see whether it is Du Val or not at this point, we calculate its Taylor series at the point. It turns out the the second term is the following:
$$60(-27x^2+42xy+4yz).$$ It is easy to see that this quadratic form has rank three, so we are lucky and the point $(0,0,-3)$ is the simplest possible singularity - an ordinary double point (of course Du Val).
Note that, in order to conclude whether this surface is $K3$ or not, we still need to check what kind of singularities it has at infinity (at three planes $x=\infty$, $y=\infty$, $z=\infty$).