In the case you are interested in there is a simple generating (Dirichlet) series:
$$ \sum_{n=1}^\infty \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}.$$
From this you can either use a convolution argument or a Perron formula type argument to get an asymptotic formula. In particular, I believe it follows that
$$\sum_{n\leq x} d(n^2) = \frac{3}{\pi^2}x \log^2 x +O(x \log x). $$
With more work, you can get lower-order terms of size $\asymp x \log x$ and $\asymp x.$

**Edit:** There seems to some disagreement on whether the coefficient of the leading order term is $\frac{3}{\pi^2}$ or $\frac{3}{2\pi^2}$. I believe that $\frac{3}{\pi^2}$ is correct. Here are three bits of reasoning:
From the generating series, we have
$$ \sum_{n\leq x}d(n^2) = \sum_{n\leq x} \sum_{k\ell^2=n} d_3(k)\mu(\ell) = \sum_{\ell\leq \sqrt{x}} \mu(\ell) \sum_{k\leq x/\ell^2} d_3(k) $$
where $d_3(n)$ denotes the number of ways to write $n$ as a product of three positive divisors and $\mu(\ell)$ is the Moebius function. By a standard estimate
$$ \sum_{k\leq x/\ell^2} d_3(k) = \frac{x}{2\ell^2}\log^2(x/\ell^2) + O\left(\frac{x\log x}{\ell^2} \right)$$
from which it follows that
$$ \sum_{n\leq x}d(n^2) = \frac{x \log^2 x}{2} \sum_{\ell \leq \sqrt{x}} \frac{\mu(\ell)}{\ell^2} +O(x\log x) = \frac{3 x}{\pi^2}\log^2 x +O(x\log x).$$
Alternatively, a Perron formula (e.g. Prime Number Theorem) type argument can be used to show that
$$ \sum_{n\leq x}d(n^2) = \text{Res}_{s=1} \frac{\zeta^3(s)}{\zeta(2s)} \frac{x^s}{s} +o(x) = \frac{3x}{\pi^2}\log^2 x + O(x\log x).$$
Moreover, in Mathematica, you can use *DivisorSigma[0, n^2]* to calculate $d(n^2)$. For $x=1,000,000$ I get that
$$ \frac{\pi^2}{3x\log^2x}\sum_{n\leq x} d(n^2) \approx 1.27305392....$$
The slow convergence to 1 is from the influence of lower-order terms. Notice, however, that the value it is not anywhere near $1/2$. However, if I define $F(x)$ to be the residue of$\frac{\zeta^3(s)}{\zeta(2s)}\frac{x^s}{s}$ at $s=1$, I get that
$$ \frac{1}{F(x)}\sum_{n\leq x} d(n^2) \approx 1.0000073....$$

not$\alpha=x$. By definition, $$d_\alpha (n)=|\{d|n:\ d\leq \alpha\}|,$$ so to be sure to get all of the divisors of $n^2$ for $n\leq x$, we must have $\alpha =x^2$. There is a way to modify Broughan's proof by using the hyperbola method so that the error depends on $\sqrt{\alpha}$. $\endgroup$ – Eric Naslund Apr 17 '12 at 16:03