一條很難的數學問題,求解答

Assume your question states that ∂f/∂x = 3x^2y+e^y, ∂f/∂y = x^3-2+xe^y

f = x^3y+xe^y+f(y), where f(y) is a function of y
f = x^3y-2y+xe^y+f(x), where f(y) is a function of x

Therefore f(x,y) = x^3y-2y+xe^y
f(2,0) = 2e^0 = 2

2013-06-01 09:44:07 補充：
Correction: f = x^3y-2y+xe^y+f(x), where f(x) is a function of x

2013-06-01 10:20:11 補充：
Correction: f(x,y) = x^3y-2y+xe^y+f0, where f0 is a constant
f(2,0) = 2+f0

f(2,0)=??????