For a precise asymptotic, we have that $$\sum_{d=1}^{N}\log d\sum_{n=1}^{N/d}\frac{\phi(n)}{\log(nd)}=-\frac{\zeta'(2)}{\zeta(2)}\text{li}(N^2)+O\left(N\right),$$ where $\text{li}(N)$ is the logarithmic Integral.
Proof:
We We may rewrite the above sum as $$\sum_{d\leq N}\log d\sum_{n\leq N,\ d|n}\frac{\phi\left(\frac{n}{d}\right)}{\log(n)},$$ and uppon switching the order this is
$$\sum_{n\leq N}\frac{1}{\log(n)}\sum_{d|n}\phi\left(\frac{n}{d}\right)\log d.$$ By using the fact that $\phi*\log=\Lambda*\text{Id},$ the above equals $$\sum_{ab\leq N}\frac{\Lambda(a)b}{\log(ab)}=\sum_{k\leq N}\frac{\left(\Lambda*\text{Id}\right)(k)}{\log k}.$$ Now, we have that
$$\sum_{ab\leq N}\Lambda(a)b=\sum_{a\leq N}\Lambda(a)\sum_{b\leq\frac{N}{a}}b=\frac{1}{2}\sum_{a\leq N}\Lambda(a)\left(\left[\frac{N}{a}\right]^{2}+\left[\frac{N}{a}\right]\right)$$
$$\frac{N^{2}}{2}\sum_{a\leq N}\frac{\Lambda(a)}{a^{2}}+O\left(N\log N\right)=-\frac{\zeta^{'}(2)}{2\zeta(2)}N^{2}+O(N\log N).$$$$=\frac{N^{2}}{2}\sum_{a\leq N}\frac{\Lambda(a)}{a^{2}}+O\left(N\log N\right)=-\frac{\zeta^{'}(2)}{2\zeta(2)}N^{2}+O(N\log N).$$
Since $$\sum_{k\leq N}\frac{\Lambda*\text{Id}(k)}{\log k}=\int_{2}^{N}\frac{1}{\log x}d\left(\sum_{k\leq x}\Lambda*\text{Id}(k)\right),$$ by applying partial summation, we are able to recover that $$\sum_{ab\leq x}\frac{\Lambda(a)b}{\log(ab)}=-\frac{\zeta'(2)}{\zeta(2)}\text{li}(N^2)+O(N).$$