I'm going to take a course about diferentiable manifolds next semester and I'm preparing for it by solving some of the problems from the book An Introduction to Differential Manifolds by D. Barden and C. Thomas. My understanding of the material is very rudimentary, I'm just memorizing the definitions and the theorems so far without reading their proofs. So, please post only the solutions that use the basic definitions and theorems.
Show that the image of $S^2 \subset \mathbb{R}^3$ under the map $f: (x,y,z) \to (x^2,y^2,z^2,\sqrt{2}yz, \sqrt{2}zx, \sqrt{2}xy)$ is a submanifold of $\mathbb{R}^6$. How does it relate to $\mathbb{R}P^2$?
Introducing charts on the $\mathrm{Im}(f)\mid_{S^2}$ seems not trivial. However, I have made an observation that I think handles the first part.
$$Df=\begin{bmatrix} 2x & 0 & 0\\ 0 & 2y & 0\\ 0 & 0 & 2z\\ 0 & \sqrt{2}z & \sqrt{2}y\\ \sqrt{2}z & 0 & \sqrt{2}x\\ \sqrt{2}y & \sqrt{2}x & 0 \end{bmatrix}$$
has constant rank $2$ over $S$ because at least one of the coordinates $x,y,z$ must be non-zero and then I can find a $2\times 2$ sub-matrix with non-zero determinant. If necessary, I can first exchange two rows to create this sub-matrix.
I think this shows that the tangent to $\mathrm{Im}(f)\mid_{S^2}$ is always a plane, so it's a $2$-dimensional manifold in $\mathbb{R}^6$. But I don't know how to make my reasoning rigorous.
I think that $\mathrm{Im}(f)$ is diffeomoprhic to $\mathbb{R}P^2$. Well, at least they have the same dimension. Plus, the image of $f$ is given by a homogeneous polynomial. I don't know how to proceed from here.
You are right, $\text{Im}(f)$ and $\mathbb{R}P^n$ are diffeomorphic.
Recall that we have a quotient map $p:S^n \to \mathbb{R}P^n$ which identifies antipodal points. Note that $f$ is constant over the fibers of $p$ so it pass to the quotient, that is, there exists a unique differentiable function $\tilde{f} : \mathbb{R}P^n \to \mathbb{R}^6$ such that $\tilde{f}\circ p = f$.
Note that $\tilde{f}$ is injective so its corestriction $g := \tilde{f}|^{\text{Im}(f)}:\mathbb{R}P^n \to \text{Im}(f)$ is a bijection. Now, because $\mathbb{R}P^n$ is compact and manifolds are Hausdorff spaces we have that $g$ is a homeomorphism and then by the Inverse Function Theorem it is a diffeomorphism.