As pointed out by Brian M. Scott, the region over which we need to integrate is the triangle with corners $(0,0)$, $(\frac{1}{4},\frac{1}{4} )$, and $(0,\frac{1}{2})$. I would like to point out a slightly different approach to the double integral that perhaps makes the integration easier.
You integrated first with respect to $y$, and then with respect to $x$. That has the disadvantage that you first get something that involves a couple of logarithms, which you then must integrate again.
Let us explore the alternative of integrating first with respect to $x$. Then the first integration will be trivial, since $1/y$ can be treated as a constant.
The downside (look at the picture!) is that for $0 \le y \le \frac{1}{4}$, $x$ goes from $0$ to $y$, while for $\frac{1}{4}\le y\le \frac{1}{2}$, $x$ goes from $0$ to $\frac{1}{2}-y$. So we will have to evaluate two integrals. We now do this, to show it is not hard.
First we evaluate $\int_{y=0}^{\frac{1}{4}}\left(\int_{x=0}^y\frac{dx}{y}\right)dy.$ Very easily, the inner integral is $1$, so our integral is $\dfrac{1}{4}$. Next we evaluate $\int_{y=\frac{1}{4}}^{\frac{1}{2}}\left(\int_{x=0}^{\frac{1}{2}-y}\frac{dx}{y}\right)dy.$ The inner integral is $\dfrac{1}{2y}-1$, so our integral is $(1/2)(\ln(1/2)-\ln(1/4)) -(1/2-1/4)$. This simplifies to $(1/2)\ln 2-1/4$. Add the two parts. We get $(1/2)\ln 2$. Finally, as in your calculation, the answer to the original question is $1-(1/2)\ln 2$.
Remark: That was easy, but in fact one can do better, by making the change of variable $w=x+y$.