The curve C has equation x^2 – 3xy – 4y^2 + 64 = 0; find dy/dx in terms of x and y, and thus find the coordinates of the points on C where dy/dx = 0

Start by differentiating x² – 3xy – 4y² + 64 = 0 with respect to x to obtain an equation in x, y and dy/dx:
2x – 3y – 3x.dy/dx – 8y.dy/dx = 0 [use the product rule to differentiate –3xy
dy/dx(3x + 8y) = 2x – 3y dy/dx = (2x – 3y)/(3x + 8y)
We want to find out what happens when dy/dx = 0. The fraction will equal zero when the numerator equals zero, so we are interested in this situation:
2x – 3y = 0 y = 2x/3
Substitute y = 2x/3 into the original equation to find where this case applies on C:
x² – 3x(2x/3) – 4(2x/3)² + 64 = 0
x² – 2x² – 16x²/9 + 64 = 0
25x²/9 = 64
x² = 576/25 x = ±24/5 [don't forget that square-rooting a number gives you both a positive and negative answer]
Now we can use the condition y = 2x/3 to find the corresponding values of y. dy/dx = 0 when (24/5, 16/5) and (–24/5, –16/5)