At which points of the surface $\displaystyle x^3y-xyz+z^2=16$ is the tangent plane horizontal?

I made a function $\displaystyle F(x,y,z)=x^3y-xyz+z^2-16$ and set the partials to x and y as zero. I found three points this way: (0,0,4), (0,0,-4). and (2,0,4). However, the answer key states that (-2,0,4) is also a point. For me, this point didn't work out due to getting an imaginary root. I would much appreciate if someone would show me how to get this last point.