Prove that \begin{equation}\sum_{n=1}^{\infty}\frac{d(n^2)}{n^s}=\frac{\zeta(s)^3}{\zeta(2s)},\end{equation}where $d(n)$ denotes the number of divisors of $n$.
My solution: Observing that we have a Dirichlet series of a multiplicative function $d(n^2)$ on the left, under some conditions (that $d(n^2)$ doesn't grow too quickly we have the Euler product representation, for $\textrm{Re}(s)>1$ that\begin{align}\sum_{n=1}^{\infty}\frac{d(n^2)}{n^s}&=\prod_p\left(\sum_{e=0}^{\infty}\frac{d(p^{2e})}{p^{es}}\right)\\&=\prod_p\left(\sum_{e=0}^{\infty}\frac{2e+1}{p^{es}}\right),\end{align} using the fact that $d(p^j)=j+1$ for prime $p$ and positive integer $j$. This is where I am getting stuck. We could try and expand this out a bit further to get \begin{equation}\prod_p\left(\sum_{e=0}^{\infty}\frac{2e}{p^{es}}+\frac{1}{p^{es}}\right)=\prod_p\left(\left(\sum_{e=0}^{\infty}\frac{2e}{p^{es}}\right)+\left(\sum_{e=0}^{\infty}\frac{1}{p^{es}}\right)\right).\end{equation} There doesn't seem to be a straightforward way to get the RHS of the desired expression from this, so I wonder if perhaps my approach is wrong.