Oh!! yeahh. I see it. You are correct.

`y=ln(((2x^3-14)^(1/3))/((1-x^2)^5))`

Let `((2x^3-14)^(1/3))/((1-x^2)^5) = f(x)`

Then,

`y = ln(f(x))`

`(dy)/(dx) = (f'(x))/(f(x))`

`f(x) = ((2x^3-14)^(1/3))/((1-x^2)^5)`

`f'(x) = ((1/3)*2*3x^2(2x^3-14)^(-2/3)(1-x^2)^5 -(5*(-1)*2x*(1-x^2)^4(2x^3-14)^(1/3)))/((1-x^2)^10)`

`f'(x) = (2x^2(2x^3-14)^(-2/3)(1-x^2) +(10x(2x^3-14)^(1/3)))/((1-x^2)^6)`

`f'(x) = (2x^2(1-x^2) +(10x(2x^3-14)))/((1-x^2)^6(2x^3-14)^(2/3))`

`f'(x) = (2x^2-2x^4+20x^4-14x)/((1-x^2)^6(2x^3-14)^(2/3))`

`f'(x) = (18x^4+2x^2-14x)/((1-x^2)^6(2x^3-14)^(2/3))`

`(dy)/(dx) = ((18x^4+2x^2-14x)/((1-x^2)^6(2x^3-14)^(2/3)))/(((2x^3-14)^(1/3))/((1-x^2)^5))`

`(dy)/(dx) = ((18x^4+2x^2-14x)/((1-x^2)(2x^3-14)^(1/3)))`

`y=ln((sqrt(3(2x^(3)-14)))/(1-x^(2))^(5))`

To ease the calculations, I am going to use a substitution as,

`(sqrt(3(2x^(3)-14)))/(1-x^(2))^(5) = f(x)`

Then, the equation changes to,

`y = ln(f(x))`

`(dy)/(dx) = (f'(x))/(f(x))`

`f(x) = (sqrt(3(2x^(3)-14)))/(1-x^(2))^(5)`

`f'(x) = ((1/2)*3*2**3x^2*(3(2x^3-14))^(-1/2)(1-x^2)^5 - sqrt(3(2x^3-14))*5*(-1)*2*x*(1-x^2)^4)/((1-x^2)^10)`

`f'(x) = (9x^2*(3(2x^3-14))^(-1/2)(1-x^2) + 10xsqrt(3(2x^3-14)))/((1-x^2)^6)`

`f'(x) = (9x^2(1-x^2) + 10x(3(2x^3-14)))/((1-x^2)^6sqrt(3(2x^3-14)))`

`f'(x) = (9x^2-9x^4+60x^4-420x)/((1-x^2)^6sqrt(3(2x^3-14)))`

`f'(x) = (51x^4+9x^2-420x)/((1-x^2)^6sqrt(3(2x^3-14)))`

Therefore,

`(dy)/(dx) = ((51x^4+9x^2-420x)/((1-x^2)^6sqrt(3(2x^3-14))))/((sqrt(3(2x^(3)-14)))/(1-x^(2))^(5))`

`(dy)/(dx) = ((51x^4+9x^2-420x)/((1-x^2)(3(2x^3-14))))`