differential topology - Checking whether the image of a smooth map is a manifold

The specific $F(M)$ is not a smooth submanifold. Here is an argument.

To simplify formulas, I renormalize the sphere: let it be the set of $(z_1,z_2)inmathbb C^2$ such that $|z_1|^2+|z_2^2|=2$ rather than 1. Then, as Gregory Arone pointed out, $F(M)$ is the set of $(b,c)inmathbb C^2$ such that the roots $z_1,z_2$ of the equation $z^2-bz+c$ satisfy $|z_1|^2+|z_2^2|=2$. I claim that it is not a smooth manifold near the point $p:=(2,1)in F(M)$.

Let us intersect $F(M)$ with two planes: ${b=2}$ and ${c=1}$. If it is is a smooth submanifold, at least one of the intersections should be locally (near $p$) a 1-dimensional smooth submanifold of the respective plane (otherwise both planes are tangent to $F(M)$ at $p$, but this is impossible since they span the whole space). But this is not the case:

If $b=2$, the equation is $z^2-2z+c=0$, hence $z_1+z_2=2$, then $|z_1|+|z_2|ge 2$ and therefore $|z_1|^2+|z_2|^2ge 2$. Equality is attained only for $z_1=z_2=1$, thus the intersection is a single point $c=1$, not a 1-dimensional submanifold.

If $c=1$, the equation is $z^2-bz+1$, hence $z_1z_2=1$, then $|z_1|cdot|z_2|=1$ and therefore $|z_1|^2+|z_2|^2ge 2$. The equality is attained if and only if $|z_1|=|z_2|=1$, so $z_1$ and $z_2$ are conjugate to each other. The set of $b$'s for which this happens is the real line segment $[-2,2]$ which is not a submanifold near 2.