= Solution
The printed product omits the node $x_0$: its lower index must be zero for the stated formula to hold. For example, with $n=1$, nodes $0,1$, and data $f_0=0,f_1=1,g_0=g_1=0$, the printed $L_{1,1}=1$ gives $h=1$, which fails the first value condition. Use the corrected <Lagrange basis polynomials>
$$
L_k(x)=\prod_{\substack{0\le i\le n\\i\ne k}}\frac{x-x_i}{x_k-x_i}.
$$
They satisfy $L_k(x_j)=\delta_{kj}$. Define $H_k=L_k^2[1-2(x-x_k)L_k'(x_k)]$ and $K_k=L_k^2(x-x_k)$. At a node with $j\ne k$, the double zero of $L_k^2$ gives $H_k(x_j)=H_k'(x_j)=K_k(x_j)=K_k'(x_j)=0$. At $x_k$, direct differentiation gives
$$
H_k(x_k)=1,\quad H_k'(x_k)=0,\quad K_k(x_k)=0,\quad K_k'(x_k)=1.
$$
Both <polynomials> have degree at most $2n+1$. Consequently $h=\sum_k(f_kH_k+g_kK_k)$ has the specified values and <derivatives>, and is the required <Hermite interpolating polynomial>. The source product is retained literally in the private transcription; this solution identifies its necessary correction.
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