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. ?
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, $$\begin{align} \sum_{y\le p < x^{1/2}} \frac{x/p}{\log (x/p)}&=x\int_y^{x^{1/2}-} f(t) d\pi(t)\\ &=x f(t)\pi(t) \Bigg\vert_{y}^{x^{1/2}-}-\int_y^{x^{1/2}-}xf'(t)\pi(t)dt + O(x/(\log x)^2)\\ &=x\left[\frac{\log \log t - \log(\log x - \log t)}{\log x}\right]_y^{x^{1/2}-}+ O(x/(\log x)^2)\\ &=\frac x{\log x} \log(u-1) + O(x/(\log x)^2). \end{align}$$