Hello, can someone help me solve this problem? I tried working it around but I can't arrive at the correct answer. Thank you!

Problem: Find the equation of the curve for which y''=12/x^{3} if it passes through (1,0) and is tangent to the line 6x+y=6 at that point.

Answer: xy+6x=6

Source: Elements of Calculus and Analytic Geometry by Reyes and Chua

$6x + y = 6$

$6 + y' = 0$

$y' = -6$

$y''=12x^{-3}$

$y' = -6x^{-2} + C_1$ → Eq (1)

$y = 6x^{-1} + C_1x + C_2$ → Eq (2)

At the point of tangency (1, 0), y' = -6

$-6 = -6 + C_1$

$C_1 = 0$

From Eq (2)

$y = 6x^{-1} + C_2$

$0 = 6 + C_2$

$C_2 = -6$

Thus,

$y = \dfrac{6}{x} - 6$ → Eq (2)

$xy = 6 - 6x$

$xy + 6x = 6$

answerGot it now, thanks! ;)