Yes, it is true. Your sum has the same asymptotics as the integral $\int_1^\infty \frac{\log x}{x}\arctan{\frac{y}x}dx$ by standard arguments (the integrated function is decreasing for $x>e$, say, and each specific summand is bounded, this is quite enough). Next, we denote $x=y/z$ to get the integral $\int_0^{y}\frac{\log(y)-\log(z)}z\arctan z dz$. We have $\int_0^{y}\frac{1}z\arctan z dz\sim \frac{\pi}2\log y$ and $\int_0^{y}\frac{\log(z)}z\arctan z dz\sim \frac{\pi}4\log^2 y$, both by l'Hospital rule.

Yes, it is true. 

Your sum has the same asymptotics as the integral $\int_1^\infty \frac{\log x}{x}\arctan{\frac{y}x}dx$ by standard arguments (the integrated function is decreasing for $x>e$, say, and each specific summand is bounded, this is quite enough).

Next, we denote $x=y/z$ to get the integral $\int_0^{y}\frac{\log(y)-\log(z)}z\arctan z dz$. We have $\int_0^{y}\frac{1}z\arctan z dz\sim \frac{\pi}2\log y$ and $\int_0^{y}\frac{\log(z)}z\arctan z dz\sim \frac{\pi}4\log^2 y$, both by L'Hôpital's rule.