Consider the closed curve in the day plane:

2x^2-2xy+y^3=14

a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k

Consider the closed curve in the xy plane:

2x^2-2xy+y^3=14

a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k

Your dy/dx should be

(2y-4x)/(3y^2 - 2x)
(the way you have it, only the -4x is divided by 3y^2 )

so when y = 2 , in the original,
2x^2 - 4x + 8 = 14
2x^2 - 4x - 6 = 0
x^2 - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3 or x = -1

b) you probably meant to say:
Find the equation of the tangent lines when y = 2

so the points of contact of the two tangents are
(3,2) and (-1,2)

at (3,2), slope = dy/dx = (4 - 8)/(12 - 6) = -4/6 = -2/3
equation of tangent:
y-2 = (-2/3)(x-3)
3y - 6 = -2x + 6
2x + 3y = 12 or y = (-2/3)x + 4

follow the same steps for the tangent to the other point

c) Not really sure what they want for that part.
Have you studied differentials?
If the point (2.5 , k) is supposed to be "close" to (3,2), that's quite a spread even if we use differentials
Just subbing in x = 2.5 yields the cubic
y^3 - 5y = -1.5

An appr just taking a few values of y gave me
y = appr 2.4 (y = 2.3732...)

To find the equation of the lines tangent to the curve, we will use the first derivative we obtained in part a) to find the slope of the tangent line at a given point.

a) We already have the derivative:
dy/dx = (2y - 4x) / (3y^2 - 2x)

b) To find the equation of the tangent line when y = 2, we need to find the corresponding x-coordinate on the curve.

Substitute y = 2 into the original equation:
2x^2 - 2x(2) + 2^3 = 14
2x^2 - 4x + 8 = 14
2x^2 - 4x - 6 = 0

Now, solve this quadratic equation for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(-4) ± √((-4)^2 - 4(2)(-6))) / (2(2))
x = (4 ± √(16 + 48)) / 4
x = (4 ± √64) / 4
x = (4 ± 8) / 4

So we have two possible x-values: x = (4 + 8) / 4 = 3 or x = (4 - 8) / 4 = -1/2

Now we have the point (x, y) on the curve when y = 2:
For x = 3, we have the point (3, 2)
For x = -1/2, we have the point (-1/2, 2)

Now, using the derivative dy/dx, we can find the slope of the tangent line at each of these points:

At (3, 2):
dy/dx = (2(2) - 4(3)) / (3(2)^2 - 2(3))
= (4 - 12) / (12 - 6)
= -8 / 6
= -4 / 3

So at (3, 2), the slope of the tangent line is -4/3.

At (-1/2, 2):
dy/dx = (2(2) - 4(-1/2)) / (3(2)^2 - 2(-1/2))
= (4 + 2) / (12 + 1)
= 6 / 13

So at (-1/2, 2), the slope of the tangent line is 6/13.

Now we have the slope of the tangent lines when y = 2. We can use the point-slope form of the equation for the lines to find their equations.

For the point (3, 2):
y - 2 = (-4/3)(x - 3)

For the point (-1/2, 2):
y - 2 = (6/13)(x + 1/2)

Therefore, the equations of the tangent lines to the curve when y = 2 are:
y = (-4/3)x + 10
y = (6/13)x + (67/13)

c) Now to find the best approximation of the value of k for the point (2.5, k), we will use the tangent line when y = 2.

Using the equation of the tangent line with slope -4/3:
y - 2 = (-4/3)(x - 3)

Substitute x = 2.5 and solve for y:
y - 2 = (-4/3)(2.5 - 3)
y - 2 = (-4/3)(-0.5)
y - 2 = 2/3
y = 2/3 + 2
y = 8/3

The best approximation of the value of k for the point (2.5, k) on the curve is k = 8/3.