$$ \sum_{k=1}^\infty \frac{k^2 q^k}{(1-q^k)^2} = \sum_{n=1}^\infty \sigma(n) n q^n$$
$$ \sum_{k=1}^\infty \frac{k q^k}{1-q^k} = \sum_{n=1}^\infty \sigma(n) q^n$$
$$ \left(\sum_{k=1}^\infty \frac{k q^k}{1-q^k}\right)^2 = \sum_{n=1}^\infty \sum_{m=1}^{n-1} \sigma(m) \sigma(n-m) q^n $$
$$ \sum_{k=1}^\infty \frac{k^3 q^k}{1-q^k} = \sum_{n=1}^\infty \sigma_3(n) q^n $$
so your identity is saying

$$ 6 n \sigma(n) + 12 \sum_{m=1}^{n-1} \sigma(m)\sigma(n-m) = 5 \sigma_3(n) + \sigma(n)$$

Hmm, surely that's got to be known. Adding number-theory to the tags.