Dirichlet hyperbola methods : estimate functions # of ordered pairs Let $f$ be an arithmetic function defined by $$f(n) = |A_n|$$ where $A_n = \{(a, b) : n = ab^2\}$.Estimate $$\sum_{n \leq x} f(n)$$ where $x \in \mathbb{R}^+$, using Dirichlet hyperbola method. The error term should be $O(x^{1/3})$.
Dirichlet hyperbola method requires the function $f$ to be written as Dirichlet convolution of two functions. The problem is I try to compute the values of $f$, and try to guess $g ,h$ such that $f = g * h$. But it is not successful.
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According to Adam Hughes answer, $$f(n) = d(n) * \lambda(n).$$
Set $$D(n) = \sum_{n \leq x} d(n), \Lambda(n) = \sum_{n \leq x} \lambda(n).$$
Let $a, b >0$ such that $ab = x,$ then by Dirichlet heperbola method
$$\sum_{n \leq x} f(n) = \sum_{c \leq a}d(c)\Lambda(x/c) + \sum_{e \leq b}\lambda(e)D(x/e) - D(a)\Lambda(b).$$
The problem is all of the examples I have seen, eg : $\sum_{n \leq x} d(n) = \sum 1*1$, know explicitly that $\sum_{n \leq x} 1 = [x].$ However, here I try to derive, and then search some information on the internet, the summation fomular for $$\Lambda(n), D(d)$$ are complicated. 
http://mathworld.wolfram.com/LiouvilleFunction.html
I try to estimate it with using big O, and follows other examples, but I cannot do it.
Any help, or hint how to do the estimation ? How to deal with that $\Lambda, D$ if the summation formula is not simple, or sometimes no formula ?
Thank you in advanced.
PS link to a new question Estimate # of order paired function using Dirichlet hyperbola method

For each $n$ we are computing $f(n) = \displaystyle\sum_{ab^2=n} 1$. Now the indicator function of the square is just $\sum_{d|n}\lambda(d)$ where here $\lambda(d)$ is the Liouville function, $\lambda(d) = (-1)^{\Omega(d)}$. Writing this out we have
$$f(n) = \sum_{d|n}\sum_{e|d}(-1)^{\Omega(e)}$$
Because this will have a $1$ in the outer sum exactly when the divisor $d$ is a perfect square and note that $a$ is totally determined by $b^2$ so we need only count the number of square factors.
As always, we reverse the order of summation using $ef = d, dg = n$ to produce
$$f(n) = \sum_{efg = n}(-1)^{\Omega(e)}\cdot 1(f)\cdot 1(g)$$
which reveals itself as a triple convolution, $1*1*\lambda$. Since we know $(1*1)(n) = d(n)$ we can write this as $d*\lambda(n)$ with $d$ the number of divisors function. Here you have something neatly written as a convolution of two basic arithmetic functions, as desired.
