Order: z=f(x, y), then:

dz=(？ f/？ x)dx+(？ f/？ y)dy

Therefore:

dz=-x？ 【e^(-x)]dx+f'y d(x？ )={2xf'y-x？ [e^(-x)]}dx

At this moment:

z=f(x，x？ )=x？ [e^(-x]

dz=d{x？ [e^(-x)]}=2x[e^(-x)]dx-x？ [e^(-x)]dx=(2x-x？ )[e^(-x)]dx

∴

2xf'y-x？ [e^(-x)]=(2x-x？ )[e^(-x)]

f'y=e^(-x)

Choose C.

To answer your question:

1, I don't understand it because you don't understand the role of independent variables, intermediate variables and dependent variables in finding partial derivatives;

2、f'x(x，x？ ) is f'x(x, y) in y=x? Instead of f(x, x? ) export values, similar to unary functions, such as:

What does f'(2) mean? Please experience it!

To sum up, your understanding of partial derivatives of binary functions is limited to formulas, but you don't understand them in essence!