I was looking for a pattern among these below:
$$ \sum_{n=1}^{\infty} \frac{\sigma_0(n^2)}{n^s} = \zeta^2(s) \sum_{n=1}^{\infty} \frac{ \mu^2(n)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)} $$ $$ \sum_{n=1}^{\infty} \frac{\sigma_0(n) \sigma_0(n^2)}{n^s} = \zeta^3(s) \sum_{n=1}^{\infty} \mu^2(n) \frac{3^{\Omega(n)}}{n^s} $$ $$ \sum_{n=1}^{\infty} \frac{\sigma^2_0(n) \sigma_0(n^2)}{n^s} = \zeta^4(s) \left ( \sum_{n=1}^{\infty} \frac{\mu^2(n)}{n^s} \left (\frac{3}{4-\sqrt{13}}\right )^{\Omega(n)} \right ) \left ( \sum_{n=1}^{\infty} \frac{\mu^2(n)}{n^s} \left (\frac{3}{4+\sqrt{13}}\right )^{\Omega(n)} \right ) \\ $$
To hopefully find $\alpha(n)$ in:
$$ \sum_{n=1}^{\infty} \frac{\sigma_0(n^2)}{\sigma_0(n)}\frac{1}{n^s} = \zeta(s) \sum_{n=1}^{\infty} \frac{ \mu^2(n) \cdot \alpha(n) }{n^s} \tag{1} $$
$(1)$ can be expanded as: $$ \prod _{n=1}^{\infty} \left(\frac{2}{1-\frac{1}{\left(p_n\right){}^s}}+\left(p_n\right){}^s \log \left(1-\frac{1}{\left(p_n\right){}^s}\right)\right) $$ but since it's not a polynomial, I can't figure a way to extract $\alpha(n)$.
Any ideas?
Edit: $\sigma_0(n)$ is the divisor function, $\mu(n)$ is the Mobius function, and $\Omega(n)$ is the prime counting function with multiplicity.
Continuing your computations, \begin{align} F(s)&:=\frac1{\zeta(s)}\sum_{n\geqslant 1}\frac{\sigma_0(n^2)}{\sigma_0(n)}\frac1{n^s}=\prod_{p\in\mathbb{P}}P(p^{-s}), \\P(z)&:=2+\frac{(1-z)\log(1-z)}{z}=1+\sum_{n\geqslant 1}\frac{z^n}{n(n+1)}. \end{align} Hence $F(s)=\sum_{n\geqslant 1}f(n)n^{-s}$ with the multiplicative function $f$ defined by $$f\left(\prod_{k=1}^n p_k^{a_k}\right)=\prod_{k=1}^n\frac1{a_k(a_k+1)}.$$
Thus $f(n)$ is not a multiple of $\mu^2(n)$. In terms of known things, $$\frac1{f(n)}=\sigma_0(n)\sigma_0\left(\frac{n}{\operatorname{rad}(n)}\right).$$