6. The curve integral has nothing to do with the path, so (? /? x)[φ(x)-x^2/2]y=(？ /? y)[(3/2)y^2φ(x)]，

[φ'(x)-x]y=3yφ(x), that is, φ'(x)-3φ(x)=x is a first-order linear differential equation, then

φ(x)=e^(∫3dx)[∫xe^(-∫3dx)dx+C]

e^(3x)[∫xe^(-3x)dx+c e^(3x)[(- 1/3)∫xde^(-3x)+c]

= e^(3x)[(- 1/3)xe^(-3x)+( 1/3)∫e^(-3x)dx+c]

= e^(3x)[(- 1/3)xe^(-3x)-( 1/9)e^(-3x)+c]

= -x/3- 1/9+Ce^(3x).

Select the integration path OA+AB, where O (0 0,0), A (1, 0), B (1, 1). then

∫& lt； L & gt[φ(x)-x^2/2]ydy+(3/2)y^2φ(x)dx =∫& lt； OA & gt+∫& lt； AB & gt

= 0+∫& lt； 0， 1 & gt； & gt[φ( 1)- 1/2]ydy =[φ( 1)- 1/2]( 1/2)= 1/4，

Get φ (1)-1/2 =1/2, φ (1) = 1. Substitute φ(x) to get.

-1/3-1/9+Ce3 =1,c = 13/(9e 3), so

φ(x)= -x/3- 1/9+ 13e^(3x-3)/9.