#2- for the curve given by 4x2+y2=48+2xy show that
dy/dx=y-4x/y-x

#3- for the curve given by 4x2+y2=48+2xy, find the positive y coordinate given that the xcoordinate is 2.

#4- for the curve given by 4x2+y2=48+2xy, show that there is a point p with x-coordinate 2 at which the line tangent to the curve at p is horizontal.

it is an implicit function, (for high school students)

1)  d(4x2+y2)/dx=d(48+2xy)/dx, after the differentiation, we obtain 
8x + 2y .dy/dx = 2y + 2x .dy/dx
8x - 2y = 2x .dy/dx - 2y .dy/dx,  4x - y = (x .- y ).dy/dx, and then 
dy/dx = y-4x / y - x
2) when x cordinate is 2
we have 4(2^2) + y^2 = 48 + 2 . 2 y, and the y^2 - 4y - 32=0, after solving the quadratic equation we find y = - 8, or y = 4, this the positive y coordinate, the answer is y = 4
3) Let be P (xp, yp) the point, since P is the point at which the line tangent to the curve at P is horizontal, so we can write 4xp2+yp2=48+2xpyp, but xp=2, we have yp^2 - 4yp - 32=0, so P(2, 4)