Show that $\sum_{n \leq x} \frac{d(n)}{\log n} = x + 2E \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right).$ 
Show that $$\sum_{n \leq x} \frac{d(n)}{\log n} = x + 2E \frac{x}{\log x} + O\left( \frac{x}{\log ^2 x} \right),$$ 
  where $d(n)$ is the number of divisors of $n$ and $E$ is Euler-Mascheroni constant.

I first showed a simpler statement since I had trouble with the denominator: $$\sum_{n \leq x} \frac{d (n)}{n} = \frac{1}{2} \log ^2 x + 2 E \log x + O(1).$$
\begin{align*}\sum_{n \leq x} \frac{d(n)}{n} & = 
\sum_{d \leq x} \frac{1}{d} \sum_{q \leq x/d} \frac{1}{q} \\ & = 
\sum_{d \leq x} \frac{1}{d} \left( \log \left (\frac{x}{d} \right) + E + O\left( \frac{d}{x} \right) \right) \\ & =
\sum_{d \leq x} \left( \frac{\log x + E}{d} - \frac{\log d}{d} + O\left( \frac{1}{x} \right) \right) \\ & =
(\log x + E) \sum_{d \leq x} \frac{1}{d} - \sum_{d \leq x} \frac{\log d}{d} + O(1) \\ & =
(\log x + E) \left( \log x + E + O\left( \frac{1}{x} \right) \right) - \left( \frac{1}{2} \log ^2 x + O(1) \right) + O(1) \\ & =
\log ^2 x + E \log x + O\left( \frac{\log x}{x} \right) + E \log x + E ^2 + O\left( \frac{1}{x} \right) - \frac{1}{2}\log ^2 x + O(1) \\ & =
\frac{1}{2} \log ^2 x + 2 E \log x + O(1).
\end{align*}
However, I am having trouble using this to show the result because of the $\log n$ in the denominator. How can we proceed and conclude the result at hand?

If we write $D(x)=\sum_{n\leq x}d(n)$ then by the Dirichlet hyperbola method we have:
$$D(x)=x\text{log}(x)+x(2\gamma -1)+\mathcal{O}(x^{1/2})$$ 
Moreover applying Abel's summation formula to your partial sum gives us:
$$\sum_{2\leq n\leq x}\frac{d(n)}{\text{log}(n)}=\frac{D(x)}{\text{log}(x)}+\int_{2}^x\frac{D(t)}{t\text{log}(t)^2}dt+\mathcal{O}\left(\frac{1}{\log(x)}\right)$$
Now note that:
$$\frac{D(x)}{\text{log}(x)}=x+2\gamma\frac{x}{\text{log}(x)}-\frac{x}{\log(x)}+\mathcal{O}\left(\frac{x^{1/2}}{\text{log}(x)}\right)$$
$$\frac{D(t)}{t\text{log}(t)^2}=\frac{1}{\text{log}(t)}+\frac{(2\gamma-1)}{\text{log}(t)^2}+\mathcal{O}\left(\frac{t^{-1/2}}{\text{log}(t)^2}\right)\implies \int_{2}^x\frac{D(t)}{t\text{log}(t)^2}=\frac{x}{\log(x)}+\mathcal{O}\left(\frac{x}{\text{log}(x)^2}\right)$$
Therefore adding these together gives us:
$$\frac{D(x)}{\text{log}(x)}+\int_{2}^x\frac{D(t)}{t\text{log}(t)^2}dt=x+2\gamma\frac{x}{\text{log}(x)}+\mathcal{O}\left(\frac{x}{\text{log}(x)^2}\right)$$
Which by our first equality proves:
$$\sum_{2\leq n\leq x}\frac{d(n)}{\text{log}(n)}=x+2\gamma\frac{x}{\text{log}(x)}+\mathcal{O}\left(\frac{x}{\text{log}(x)^2}\right)$$
