I found part B, but stuck on part A.

Question

Use implicit differentiation to find the points where the parabola defined by x^2-2xy+y^2+4x-8y+24= 0 has horizontal and vertical tangent lines.

A) The parabola has horizontal tangent lines at the point(s):

B)The parabola has vertical tangent lines at the point(s): (2,6)

The derivative of the function given is (x-y+2)/(x-y+4).
To find the horizontal tangent line you should set (x-y+2)/(x-y+4)= 0 ,but when I set it to 0 the x and y cancel each other out...

Answer 1

I agree with your derivative, and I a agree with your idea of
(x-y+2)/(x-y+4)= 0 for the horizontal tangent.

Now consider where the zero could come from.
It certainly can't come from the denominator, since we can't divide by zero (this will be considered in the next paragraph)
So it must be x-y + 2 = 0
or y = x+2
sub that back into the original x^2-2xy+y^2+4x-8y+24= 0
x^2 - 2x(x+2) + (x+2)^2 + 4x - 8(x+2) + 24= 0
4x + 12 = 0 , check my algebra.
x = -3
y = -1
So you have a horizontal tangent at (-3,-1)

for the vertical tangent, the slope must be undefined.
This can only happen if the denominator is zero.

so set x-y+4 = 0 ---> y = x+4

and proceed like did for the horizontal

Answer 2

x^2-2xy+y^2+4x-8y+24=0
2x-2y-2xy'+2yy'+4-8y'=0
y' = -(2x-2y+4)/(-2x+2y-8) = (x-y+2)/(x-y+4)

so far, so good. For y'=0, we have
y=x+2
x^2-2x(x+2)+(x+2)^2+4x-8(x+2)+24=0
x=3 so, y=5
Looks like you missed a minus sign somewhere there. A peek at the graph at

http://www.wolframalpha.com/input/?i=x%5E2-2xy%2By%5E2%2B4x-8y%2B24%3D+0

confirms this, with the horizontal tangent at (3,5).

I'll let you work out the algebra, but it appears that the vertical tangent is at (2,6)