z²=1-x²-y².
Define g(x,y)=2x²-4xy-y²+2y√(1-x²-y²)+3(1-x²-y²),
g(x,y)=2x²-4xy-y²+2y√(1-x²-y²)+3-3x²-3y²,
g(x,y)=-(x²+4xy+4y²)+2y√(1-x²-y²)+3,
g(x,y)=-(x+2y)²+2y√(1-x²-y²)+3.
When x=-2y, g(-2y,y)=2y√(1-5y²)+3 and this is a maximum, because the minimum value of (x+2y)² is zero to be subtracted from a positive value.
max g=2y√(1-5y²)+3, so we need to analyse this expression to discover its maximum value.
Let u=2y√(1-5y²)+3,
du/dy=2√(1-5y²)+(2y)(½)(-10y)/√(1-5y²),
du/dy=(2(1-5y²)-10y²)/√(1-5y²).
When 2(1-5y²)-10y²=0, 1-5y²=5y², y=±√0.1. For maximum value y=√0.1.
So max g is g(-2√0.1,√0.1)=3+1/√5=3.4472 approx.
This corresponds to f(-2√0.1,√0.1,√0.5). Note that x²+y²+z²=1.
If we take y=-√0.1 we get the minimum value.
Min g is g(2√0.1,-√0.1)=3-1/√5=2.5528 approx.
This is a representation of the surface g(x,y) and the maximum and minimum points are shown as A and B on the top surface. To view in 3D use blue/green (right eye), red (left eye) filtering glasses. Acknowledgements to the GeoGebra app for using their facilities.