φ(x^2,e^y,z)=0: 2 x φ'1+cosxe^yφ‘2+z'φ'3=0 =>z'=-[2 x φ'1+cosxe^(sinx)φ‘2] / φ'3 du/dx=f'1+cosx f'2+z' f'3 =f'1+cosx f'2-{[2 x φ'1+cosxe^(sinx)φ‘2] / φ'3} f'3