Please check my answers and let me know if I did something wrong. Thank you!

Find the partial derivative of x, and the partial derivative of y, then the partial derivative of x (1,-1), and the partal derivative of y (1,-1).

f(x,y) = x^4 y^2 -x

here is what I got
@f/@x = 3x^3 y^2 -1
@f / @y = x^4 2y
@f / @x = 3(1)^3 (-1)^2 -1 = 2
@f / @y = (1)^4 2(-1) = -2

f(x,y) = e^(2x+y)
@f/@x = 2e^(2x+y)
@f / @y = e^(y+2x)
@f / @x = 2e
@f / @y = e

Locate and classify the critical points the functions.
g(x,y) = -x^2 -2xy + y^2 + x -4y
g(x) = -2x -2y +1
g(y) = -2x + 2y -4
set them equal to zero?
g(x) = -2x -2y +1 = 0
g(y) = -2x + 2y -4 = 0
Here is what I got.
(-3/4, 5/4)

f(x,y) xe^y
f(x) = e^y
f(y) = e^(y) x
x=0

Looks good, except for the last one. It has no critical points, since Fx is never zero.

Note that the critical points are where both Fx and Fy are zero.

Let's go through each question one by one.

1. Finding partial derivatives of f(x, y) = x^4y^2 - x:
To find the partial derivative with respect to x, we treat y as a constant and differentiate x. So,
∂f/∂x = (4x^3)y^2 - 1
To find the partial derivative with respect to y, we treat x as a constant and differentiate y. So,
∂f/∂y = x^4(2y)

Now, to evaluate the partial derivatives at the point (1, -1), substitute x = 1 and y = -1 into the partial derivatives.

∂f/∂x = (4(1)^3)(-1)^2 - 1 = 4 - 1 = 3
∂f/∂y = (1)^4(2(-1)) = -2

So, the values of the partial derivatives at (1, -1) are:
∂f/∂x = 3
∂f/∂y = -2

2. Evaluating the partial derivatives of f(x, y) = e^(2x+y):
To find the partial derivative with respect to x, we treat y as a constant and differentiate e^(2x+y) with respect to x. So,
∂f/∂x = 2e^(2x+y)
To find the partial derivative with respect to y, we treat x as a constant and differentiate e^(2x+y) with respect to y. So,
∂f/∂y = e^(2x+y)

The partial derivatives are not dependent on any specific point, so there is no need to substitute any values.

3. Locating and classifying the critical points of g(x, y) = -x^2 - 2xy + y^2 + x - 4y:
To find critical points, we need to find where the partial derivatives of g(x, y) are equal to zero.

∂g/∂x = -2x - 2y + 1 = 0
∂g/∂y = -2x + 2y - 4 = 0

To solve these equations simultaneously, we can use substitution or elimination method. Solving these equations, we get x = -3/4 and y = 5/4. So, the critical point is (-3/4, 5/4).

4. Evaluating f(x, y) = xe^y at x = 0:
Substitute x = 0 into f(x, y) to evaluate it. So,
f(0, y) = 0e^y = 0

Therefore, when x = 0, f(x, y) evaluates to 0.