I assume you differentiating with respect to x ?
3x^2 = y(-sinx) + dy/dx(cosx)
dy/dx = 3x^2 + ysinx
d(dy/dx)/dx = 6x + dy/dx sinx + ycosx
= 6x + (3x^2 + ysinx)sinx + ycosx
= 6x^2 + 3x^2sinx + ysin^2 x + ycosx
check my work, should have written it on paper first
Find second differentiation for function x^3=ycosx
4 answers
or
y = x^3/cosx
y' = (cosx(3x^2) - x^3(-sinx))/cos^2x
= (3x^2 cosx + x^3 sinx)/cos^2x
now do it again
y = x^3/cosx
y' = (cosx(3x^2) - x^3(-sinx))/cos^2x
= (3x^2 cosx + x^3 sinx)/cos^2x
now do it again
Why 6x become power of 2 at the end?
That was clearly a typo, good for you to catch it