The below identity I have found experimentally.
Question. Is this true? If so, may you provide a "slick" (or any) proof. $$6\sum_{k=1}^{\infty}\frac{k^2q^k}{(1-q^k)^2}+12\left(\sum_{k=1}^{\infty}\frac{kq^k}{1-q^k}\right)^2=\sum_{k=1}^{\infty}\frac{(5k^3+k)q^k}{1-q^k}.$$
Follow the comments of Lucia and note that $$\sum_{n\ge 1}\frac{n^2q^n}{(1-q^n)^2}=q\frac{\,d}{\,dq}\sum_{n\ge 1}\frac{nq^n}{1-q^n}.$$ I believe the identity actually is the well known $$q\frac{\,d}{\,dq}L=\frac{L^2-M}{12},$$ where $$L=1-24\sum_{n\ge 1}\frac{nq^n}{1-q^n}\;\mbox{ and}\; M=1+240\sum_{n\ge 1}\frac{n^3q^n}{1-q^n}.$$ You can find it in here https://en.wikipedia.org/wiki/Eisenstein_series.
$$ \sum_{k=1}^\infty \frac{k^2 q^k}{(1-q^k)^2} = \sum_{n=1}^\infty \sigma(n) n q^n$$ $$ \sum_{k=1}^\infty \frac{k q^k}{1-q^k} = \sum_{n=1}^\infty \sigma(n) q^n$$ $$ \left(\sum_{k=1}^\infty \frac{k q^k}{1-q^k}\right)^2 = \sum_{n=1}^\infty \sum_{m=1}^{n-1} \sigma(m) \sigma(n-m) q^n $$ $$ \sum_{k=1}^\infty \frac{k^3 q^k}{1-q^k} = \sum_{n=1}^\infty \sigma_3(n) q^n $$ 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.
