This answer gives a Tauberian argument:
As the Greg Martin suggests in the comment section, the Dirichlet series
$$
F(s)=\sum_{n=1}^\infty{\mu^2(n)\varphi(n)\over n^s}=s\int_1^\infty{A(x)\over x^{s+1}}\mathrm dx\tag1
$$
has a simple pole at $s=2$ with residue
$$
K=\prod_p\left(1+{p-1\over p^2}\right)\left(1-\frac1p\right)
$$
which implies
$$
G(s)={F(s)\over s}-{K/2\over s-2}
$$
is analytic in a neighborhood containing $s=2$. Plugging in (1), we have
$$
\begin{aligned}
G(s)
&=\int_1^\infty{A(x)-Kx^2/2\over x^{s+1}}\mathrm dx \\
&=\int_1^\infty{A(\sqrt t)-Kt/2\over2t^{s/2+1}}\mathrm dt
\end{aligned}
$$
Now, Wiener-Ikehara Tauberian theorem guarantees that the RHS converges at $s=2$, so we know that
$$
\int_1^\infty{2A(\sqrt t)/K-t\over t^2}\mathrm dt
$$
converges. Now, for convenience, we set $B(x)=2A(\sqrt x)/K$, so the above formula becomes
$$
\int_1^\infty{B(t)-t\over t^2}\mathrm dt\tag2
$$
from this, we will show that $B(t)\sim t$ therefore $A(x)\sim Kx^2/2$:
If $B(t)\nsim t$ then there exists a constant $\lambda>1$ such that there exists a strictly increasing sequence $t_n$ such that $B(t_n)\ge\lambda t_n$ for all $n$, meaning
$$
\begin{aligned}
\int_{t_n}^{\lambda t_n}{B(t)-t\over t^2}\mathrm dt
&\ge\int_{t_n}^{\lambda t_n}{\lambda t_n-t\over t^2}\mathrm dt
=\int_{t_n}^{\lambda t_n}{\lambda-t/t_n\over(t/t_n)^2}\mathrm d(t/t_n) \\
&=\int_1^\lambda{\lambda-u\over u^2}\mathrm du>0
\end{aligned}
$$
which violates Cauchy's criterion for convergence of (2), so we have
$$
\limsup_{t\to\infty}{B(t)\over t}\le1
$$
Similar argument can be established for $\liminf_{t\to\infty}B(t)/t\ge1$. Consequently, we see that $B(t)\sim t$, indicating
$$
\lim_{x\to\infty}{A(x)\over x^2}=\frac K2=\frac12\prod_p\left(1+{p-1\over p^2}\right)\left(1-\frac1p\right)
$$
