I was just wondering if somebody could check over my math if they have a second.

Find $\displaystyle x^3+y^3 = 126$ and find y''(x) @ (5 , 1)

Took d/dx $\displaystyle 3x^2+3y^2(dy/dx) = 0$

Solved for dy/dx by subtracting and dividing $\displaystyle (dy/dx) = (-3x^2)/(3y^2) $

Took $\displaystyle d^2y/dx^2 $ using quotient rule $\displaystyle \frac {(-3x^2*6y(dy/dx)-3y^2(-6x))}{(3y^2)^2} $

Plugged in dy/dx to the second derivative $\displaystyle \frac {(-3x^2*6y(-3x^2/3y^2)-3y^2(-6x))}{(3y^2)^2} $ At this point does the procedure look correct? I plug in x = 5 and y = 1 and get 1260.