As y is always positive, let us see what happens in these cases
1. x>0, in this case we are subtracting a positive term from y^2 and hence possibility of minimum
2. x<0, in this case we are subtracting a negative term from y^2 and hence a possibility for maximum.
So we say g(x) = max|y^2 - xy|, x>0 for g(x) = minimum.
Now say y lies in (0,1] ie y is not equal to zero. In this case there will always exist a positive x where g(x) is always > 0.
And hence the minimum value will be possibile for x=y, ie the minimum value will be 0.
In case y=0, then g(x) = 0 for all x, and in this case also minimum value is 0.
Hence minimum value of g(x) is clearly 0.
Hope it helps.
Ashwin (IIT Madras).