e^(z)=xyz e^(z)-xyz=0 F(x,y,z)=e^(z)-xyz=0 According to formula 7 Suppose that an equation F(x,y,z)=0 defines z Implicitly in terms of x and y Then (∂ z)/(∂ x)=- (F_(x))/(F_(z)) (∂ z)/(dy)=- (F_(y))/(F_(z)) F_(x)=-yz F_(y)=-xz F_(z)=e^(z)-xy Therefore (∂ z)/(dx)=- (F_(x))/(F_(z))=- (-yz)/(e^(z)-xy)= (yz)/(e^(z)-xy) (∂ z)/(dy)=- (F_(y))/(F_(z))=- (-xz)/(e^(z)-xy)= (xz)/(e^(z)-xy) Result: (∂ z)/(dx)= (yz)/(e^(z)-xy) (∂ z)/(dy)= (xz)/(e^(z)-xy)