f(1)=-1
\(y=-x^2\), and \(y=1\)
\(1=-x^2\), solve for x, \(x=-1\), check it fits piecewise, and it does.
therefore \(f(1)=-1\)
You can also graph the equations \(y=-x^2\) with \(x \geq0\) and \(y=x+8\) with \(x<0\), go to where \(y=1\) and see what \(x\) is.