= Solution
Rescale the ambient coordinates by $q_i=a_i r_i$. For each $i\in\{1,\ldots,n+1\}$ and $\varepsilon\in\{1,-1\}$, take the relatively open set $U_i^\varepsilon$ where $\varepsilon q_i>0$. Define a <manifold chart> by retaining all the $q_j$ except $q_i$:
$$
\phi_i^\varepsilon(r)=(q_1,\ldots,\widehat{q_i},\ldots,q_{n+1})\in B^n.
$$
Here $B^n$ is the open unit <open unit ball>. The inverse of this <manifold chart> is explicit:
$$
r_j=\frac{x_j}{a_j}\quad(j\ne i),\qquad r_i=\frac{\varepsilon}{a_i}\sqrt{1-\sum_{j\ne i}x_j^2}.
$$
The coordinate labels in $x$ retain their original indices. All the $a_i$ are nonzero, so this inverse exists; the strictly positive radicand makes it smooth on the whole open <open unit ball>. Projection and this inverse are continuous, so each <manifold chart> is a <homeomorphism> onto an open subset of $\mathbb R^n$.
These $2(n+1)$ <manifold charts> cover the <ellipsoid>, because at every point at least one $q_i$ is nonzero. On an overlap with a different chart $U_j^\eta$, the coordinates of $\phi_j^\eta\circ(\phi_i^\varepsilon)^{-1}$ are the retained $x_k$ for $k\ne i,j$ together with
$$
q_i=\varepsilon\sqrt{1-\sum_{k\ne i}x_k^2}.
$$
Its domain is the open subset $\eta x_j>0$ of $B^n$. The displayed functions, and the reverse transition functions obtained by exchanging $i,j$, are smooth. Transitions between identical charts are the identity; opposite-sign charts with the same omitted coordinate do not overlap. The ambient subspace topology is <Hausdorff> and <second countable>, as it is inherited from Euclidean space. Thus this <coordinate-projection atlas of an ellipsoid> is a <smooth atlas>, proving \b[the ellipsoid is a smooth manifold of dimension $n$]. The rescaling $r\mapsto q$ also gives a <diffeomorphism> with the unit <sphere> $S^n$.
For the <three-sphere>, use its identification with <SU(2)> rather than separate local constructions. With the <Pauli matrices> $\tau_i$, put $T_i=-i\tau_i$. These form a real basis of the <SU(2) Lie algebra> and satisfy
$$
T_iT_j=-\delta_{ij}I+\epsilon_{ijk}T_k.
$$
For $q_0^2+q_1^2+q_2^2+q_3^2=1$, the identification <SU(2) as the three-sphere> is
$$
U(q)=q_0I+q_iT_i=\begin{pmatrix}q_0-iq_3&-q_2-iq_1\\q_2-iq_1&q_0+iq_3\end{pmatrix}.
$$
The matrix is unitary with determinant one. Conversely, every <SU(2) matrix> has this form, and both the correspondence and its inverse are smooth.
The left translations $L_g(h)=gh$ are <diffeomorphisms>. Define three <left-invariant vector fields> by
$$
\boxed{X_i(g)=(dL_g)_eT_i=gT_i,\qquad i=1,2,3.}
$$
Matrix multiplication makes these <vector fields> smooth globally. Since $(dL_g)_e$ is an isomorphism from the <SU(2) Lie algebra> to $T_g\mathrm{SU}(2)$, the three vectors are linearly independent at every point, and each is nowhere zero. This is the <parallelization of a Lie group by left translations>.
Transporting the <left-invariant vector fields> to the <three-sphere> gives the <quaternionic left-invariant frame on the three-sphere>:
$$
\begin{aligned}
X_1(q)&=(-q_1,q_0,q_3,-q_2),\\
X_2(q)&=(-q_2,-q_3,q_0,q_1),\\
X_3(q)&=(-q_3,q_2,-q_1,q_0).
\end{aligned}
$$
Directly, $q\cdot X_i=0$ and $X_i\cdot X_j=\delta_{ij}$ on the unit <three-sphere>. Thus the vectors are tangent and form a global orthonormal <frame of a vector bundle>. \b[The three-sphere is a parallelizable manifold], with <tangent bundle> $TS^3\cong S^3\times\mathbb R^3$.
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