My textbook is listing solutions to this equation as $2t=\pm \frac{\pi}{4}$ and $2t=\pm \frac{5\pi}{4}$ however this doesn't seem correct at all, I believe the only solutions should be $2t=\frac{\pi}{4}$ and $2t=\frac{5\pi}{4}$ since the other solutions will result in $\tan(2t)=-1$
In case it's of any importance this came from parameterizing $x^2+y^2<1$ with $x=\cos t$ and $y=\sin t$ and then plugging these into $f(x,y)=xy-y^2$ to obtain $g(t)=\frac{1}{2}(\sin 2t+\cos 2t+1)$ and then attempting to solve for when max/min occur.
Yes you are right but the better solution is $2t=\pi/4+k\pi, k\in \mathbb{Z}. $