The below identity I have found experimentally.



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> **Question.** Is this true? If so, may you provide a "slick" (or any) proof.
>  $$6\sum\_{k=1}^{\infty}\frac{k^2q^k}{(1-q^k)^2}+12\left(\sum\_{k=1}^{\infty}\frac{kq^k}{1-q^k}\right)^2=\sum\_{k=1}^{\infty}\frac{(5k^3+k)q^k}{1-q^k}.$$
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Searching for a proof for a series identity
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>.


