= Solution
Put $n=x/R$ and $q=r(R)/R$. Differentiating $y=qx$ uses $\nabla R=n$ and $\nabla q=(r'-q)n/R$. Hence
$$
\boxed{F=qI+(r'-q)n\otimes n.}
$$
Its <eigenvalue>[radial eigenvalue] is $r'$ and its two tangential <eigenvalues> are $q$. They are positive by hypothesis, so they are also the principal stretches:
$$
\boxed{v_1=r',\qquad v_2=v_3=q.}
$$
For an <isotropic> <strain energy density>[stored energy] the <nominal stress> has the same radial/tangential <projectors>,
$$
P=DW(F)=\Phi_1 n\otimes n+\Phi_2(I-n\otimes n),
$$
where symmetry of $\Phi$ gives equal tangential derivatives. Since $\operatorname{Div}(n\otimes n)=2n/R$ and the scalar coefficients depend only on $R$,
$$
\operatorname{Div}P=\left[\frac{d\Phi_1}{dR}+\frac2R(\Phi_1-\Phi_2)\right]n.
$$
Thus <equilibrium> is exactly $\boxed{d\Phi_1/dR=2(\Phi_2-\Phi_1)/R}$.
For the particular energy define $\delta=r'q^2=\det F>0$. Its two derivatives are
$$
\Phi_1=\mu+h'(\delta)q^2,\qquad\Phi_2=\mu+h'(\delta)r'q.
$$
Also $q'=(r'-q)/R$. Differentiating the first formula and substituting the <equilibrium> equation gives
$$
h''(\delta)\delta'q^2+2h'(\delta)qq'=2h'(\delta)q(r'-q)/R=2h'(\delta)qq'.
$$
The last terms cancel. Strict positivity of $h''$ and $q$ yields $\boxed{\delta'=0}$, so $r^2r'=D R^2$ for one positive constant $D$. Integration gives $r^3=DR^3+C$.
The displacement data mean $r(1)=\lambda_1$ and $r(2)=2\lambda_2$, not $r(2)=\lambda_2$. Solving the two scalar equations gives
$$
\boxed{D=\frac{8\lambda_2^3-\lambda_1^3}{7},\qquad C=\frac{8(\lambda_1^3-\lambda_2^3)}7,\qquad r=(DR^3+C)^{1/3}.}
$$
The given inequality makes $D>0$. Since $r^3$ increases from $\lambda_1^3>0$ to $8\lambda_2^3$, it is positive on the whole shell, and $r'=DR^2/r^2>0$. Thus the formula is <smooth> and <admissible> and satisfies <equilibrium>. Conversely every <admissible> radial <equilibrium> had to have this constant <determinant> and these two integration constants, proving <uniqueness>. This is a <constant-Jacobian radial equilibrium for a stretch-sum energy>.
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