Numbers divisible by the square of their largest prime factor Let $p(n)$ be greatest prime factor of $n$, denote $A=\{n\mid p^2(n)\mid n,n\in \mathbb N\}.$
$A=\{4,8,9,16,18,25,27,32,36,49,50,\cdots\},$ see also A070003.
Define $f(x)=\sum_{\substack{n\leq x\\n\in A}}1.$ Erdős proved that
$$f(x)=x \cdot e^{-(1 + o(1))\sqrt{\log x \log \log x}}.\tag1$$
Hence $f(x)=o(x),$
$$\lim_{x\to \infty}\frac{f(x)}x=0,\tag2$$
since $(2)$ is  easier than $(1)$, can you prove $(2)$ without use $(1)$?

What follows is self contained proof that $$f(x)=\sum_{\begin{array}{c}
n\leq x\\
n\in A
\end{array}}1 \ll x e^{-c\sqrt{\log x}}.$$ If you work more carefully with the friable integers, you can recover Erdős' result.  

Proof. We may write $$\sum_{\begin{array}{c}
n\leq x\\
n\in A
\end{array}}1=\sum_{\begin{array}{c}
np^{2}\leq x\\
n\in S(p)
\end{array}}1 = \sum_{p\leq\sqrt{x}}\sum_{\begin{array}{c}
n\leq x/p^{2}\\
n\in S(p)
\end{array}}1$$ where $S(y)=\left\{ n:\ P(n)\leq y\right\}$ is the set of $y$-friable integers. Notice that $$ \sum_{p\leq\sqrt{x}}\sum_{\begin{array}{c}
n\leq x/p^{2}\\
n\in S(p)
\end{array}}1 
 \leq x\sum_{B\leq p}\frac{1}{p^{2}}+\sum_{\begin{array}{c}
n\leq x\\
P(n)\leq B
\end{array}}1.\ \ \ \ \ \ \ \ \ \ (1)$$ The first term is $\leq\frac{x}{B}$, and the second term may be bounded using Rankin's Trick, which we will now use. Let $\sigma>0$, and notice that $$\sum_{n:\ P(n)\leq B}1\leq\sum_{\begin{array}{c}
n=1\\
P(n)\leq B
\end{array}}^{\infty}\left(\frac{x}{n}\right)^{\sigma}=x^{\sigma}\prod_{p}\left(1-\frac{1}{p^{\sigma}}\right)^{-1}.$$ For $\sigma=1-\frac{1}{\log B},$ the above is 
  $$\ll xe^{-\frac{\log x}{\log y}}\exp\left(e\sum_{p\leq B}\frac{1}{p}\right)\ll xe^{-\frac{\log x}{\log B}}\left(\log B\right)^{e}.$$ 
  Setting $B=e^{\sqrt{\log x}},$ it follows that $$\sum_{\begin{array}{c}
n\leq x\\
n\in A
\end{array}}1\ll xe^{-\sqrt{\log x}},$$ 
  and the result is proven.

Remark: Erdős' result may be recovered in the following manner: First, to obtain the extra $\sqrt{\log \log x}$ in the upper bound, you need a bound of the form $$\Psi(x,y)=\sum_{\begin{array}{c}
n\leq x\\
P(n)\leq y
\end{array}}1 \ll x u^{-(1+o(1))u}$$ where $u=\frac{\log x}{\log y},$ rather than Rankin's trick. Now, the asymptotic can be obtain by bounding the second term in equation $(1)$, and noticing that the first term will provide us with our main term.  
