= Solution
The <conjugate partition> $\lambda'$ has the same hook lengths as $\lambda$ and the opposite contents. Applying part d(i) to both diagrams and adding yields
$$
\sum_i\bigl(\lambda_i^2+(\lambda'_i)^2\bigr)
=2\sum_{(i,j)\in\lambda}h_{i,j}.
$$
Now sum the first identity of part d(i) over all partitions $\mu\vdash n$. Conjugation is a bijection on those partitions, so the total of $\sum_i(\mu'_i)^2$ equals the total of $\sum_i\mu_i^2$. Dividing the summed displayed identity by two gives
$$
\sum_{\mu\vdash n}\sum_i\mu_i^2
=\sum_{\mu\vdash n}\sum_{(i,j)\in\mu}h_{i,j}(\mu).
$$
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