Relation between Buchstab identity and prime number theorem 
Using the estimate $$\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{\log^2 x}\right)$$ prove that, for $x^{1/3}<y\le x^{1/2}$ and for $u=\log x/\log y$,
  $$\Phi(x,y)=\frac{x}{\log x}\{1+\log (u-1)\}+O\left(\frac{x}{\log^2 x}\right)$$
  where $\Phi(x,y)$ denotes the number of positive integers $\le x$ whose all prime factors are $> y$.

What I tried: Using Buchstab's Identity we have, $$\Phi(x,y)=\Phi(x,x^{1/2})+\sum_{x^{1/3}<p\le  x^{1/2}}\Phi(x/p,p)$$
Also we know that for $y>\sqrt x$, $\Phi(x,y)=\pi(x)-\pi(y)+1$.
Then, $$\Phi(x,y)=\frac{x}{\log x}-\frac{2x^{1/2}}{\log x}+1+\sum_{x^{1/3}<p\le  x^{1/2}}\Phi(x/p,p)$$
Then how can I formulate this? Maybe I am going in some wrong direction?
Any hint. ?
 A: The definition of $\Phi(x,y)$ is the number of positive integers $n\leq x$ whose prime divisors are $\ge y$. Also, Buchstab's identity was incorrect in the question. It should begin with
$$\begin{align}
\Phi(x,y)&=\Phi(x,x^{1/2})+\sum_{y\le p<  x^{1/2}}\Phi(x/p,p)\\
&=\pi(x)+\sum_{y\le p<x^{1/2}} (\pi(x/p)-\pi(p)) +O(x/(\log x)^2)\\
&=\frac x{\log x}+\sum_{y\le p<x^{1/2}} \left(\frac{x/p} {\log(x/p)}-\frac p{\log p}\right)+O(x/(\log x)^2)\\
&=\frac x{\log x}+\sum_{y\le p<x^{1/2}} \frac{x/p} {\log(x/p)}+O(x/(\log x)^2)\\
&=\frac x{\log x} \left(1+\log(u-1)\right) + O(x/(\log x)^2).
\end{align}$$
The sum over $p$ is treated by partial summation with $f(t)=\frac  1{t\log(x/t)}$. 
For details, 
$$\begin{align}
\sum_{y\le p < x^{1/2}} \frac{x/p}{\log (x/p)}&=x\int_y^{x^{1/2}-} f(t) d\pi(t)\\
&=x f(t)\pi(t) \Bigg\vert_{y}^{x^{1/2}-}-\int_y^{x^{1/2}-}xf'(t)\pi(t)dt + O(x/(\log x)^2)\\
&=x\left[\frac{\log \log t - \log(\log x - \log t)}{\log x}\right]_y^{x^{1/2}-}+ O(x/(\log x)^2)\\
&=\frac x{\log x} \log(u-1) + O(x/(\log x)^2).
\end{align}$$
