Let $f(x,y)=\frac{\log x}{x-y}$. Since the principal value does not seem well-defined here, let's integrate $f(x,y)+f(y,x)= \frac{\log (x/y)}{x-y}$ (which is positive) on the triangle defined by $0. Now $\int_0^x \frac{\log (x/y)}{x-y} dy = -\int_0^1 \frac{\log u}{1-u} du$ (change of variable $y=xu$), which does not depend on $x$, and so this is equal to the first integral (integrating for $x$ between $0$ and $1$).
Now $-\int_0^{1-\epsilon} \frac{\log u}{1-u} du = \int_0^{1-\epsilon} \sum_{ n \geq 1} \frac{u^{n-1}}{n} du = \sum_{n \geq 1} \frac{(1-\epsilon)^n}{n^2}$ and letting $\epsilon$ go to $0$ gives $\zeta(2)=\frac{\pi^2}{6}$.