Take the first derivative of f(x, y) with respect to x and y respectively, and get f'x and f'y; let f'x=0 f'y=0, and get a binary system of equations, and solve all stagnation points (x0, y0) (maybe more than one).

Find the second derivative of f(x, y) and get f'xx f'yy and f'xy, so that a = f' xx (x0, y0) b = f' xy (x0, y0) c = f' YY (x0, y0).

When AC-B 2 > 0, A>0, then f(x0, y0) is the minimum value, a.

When AC-B 2

When AC-B 2 = 0, it is uncertain whether there is an extreme value, and it needs to be judged again.

f'x=e^(x/2)+ 1/2*(x+y^2)e^(x/2)=[ 1+(x+y^2)/2]e^(x/2)=0

f'y=2ye^(x/2)=0

The stagnation point x0=-2 y0=0 is obtained.

f'xx= 1/2*e^(x/2)+ 1/2*[ 1+(x+y^2)/2]e^(x/2)= 1/2*e^(x/2)*[2+(x+y^2)/2]

f'xy=ye^(x/2)

f'yy=2e^(x/2)

A=f'xx(x0，y0)= 1/2e

B=0

C=2/e

ac-b^2= 1/e^2>； 0, so the original function has an extreme value.

Because A>0, f (-2,0) =-2/e is the minimum value.