= Solution
The <Inada conditions> give $U'(0+)=\infty$ and $U'(\infty)=0$. <Differentiability> and <strict concavity> make $U'$ continuous and strictly decreasing, with range $(0,\infty)$. Hence the <inverse marginal utility> $I=(U')^{-1}$ exists. The unique maximizer in the <utility conjugate> is $x=I(y)$, and
$$
\widehat U(y)=U(I(y))-yI(y),\qquad\boxed{\widehat U'(y)=-I(y).}
$$
The <derivative> formula does not require <differentiability> of $I$: compare the optimizing values at $y$ and $y+h$ to squeeze the <difference quotient> between $-I(y)$ and $-I(y+h)$, and use <continuity> of $I$. Thus the dual is continuously differentiable, strictly decreasing, and <strictly convex>, since $-I$ is strictly increasing.
For the requested second-derivative assertion, a curvature hypothesis is missing. Under the intended nondegeneracy $U''(x)<0$ for every $x>0$, inverse differentiation gives
$$
\boxed{\widehat U''(y)=-\frac1{U''(I(y))}>0,\qquad U''(x)\widehat U''(U'(x))=-1.}
$$
This proves the intended <dual differentiability with nonvanishing utility curvature>. If $U$ is only twice differentiable, it gives pointwise twice <differentiability> of the dual; continuous second <derivatives> additionally follow when $U\in C^2$.
Literal <strict concavity> does not imply nonvanishing curvature. An explicit <Inada utility with vanishing curvature> is
$$
U(x)=\log x+\frac1x-\frac1{6x^2},\quad U'(x)=\frac1x-\frac1{x^2}+\frac1{3x^3},\quad U''(x)=-\frac{(x-1)^2}{x^4}.
$$
Its <derivative> is positive, strictly decreasing, tends to infinity at zero and to zero at infinity. It is smooth and <strictly concave>, but $U''(1)=0$. At $y_0=U'(1)=1/3$, a finite <derivative> of $I$ would contradict differentiation of $U'(I(y))=y$, giving $0\cdot I'(y_0)=1$. Hence $\widehat U'$ is not differentiable there. \b[As printed, the twice-differentiable-dual claim is false; it is valid with $U''<0$.] The remaining <differentiability>, monotonicity and strict-convexity conclusions above hold under the printed hypotheses.
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