Find $\displaystyle \frac{dy}{dx}$ given that $\displaystyle y=\frac{u-17}{u+17}$ , $\displaystyle u=x^3$.

I began trying to solve it like this: $\displaystyle \frac{dy}{du}$ = $\displaystyle \frac{(u+17)(-17)-(u-17)(17)}{(u+17)^2}$ = $\displaystyle \frac{-34u}{(u+17)^2}$

$\displaystyle \frac{du}{dx}$ = $\displaystyle 3x^2$

$\displaystyle \frac{dy}{dx}$ = $\displaystyle \frac{-34x^3}{(x^3+17)^2} (3x^2)$ = $\displaystyle \frac{-102x^5}{(x^3+17)^2}$

Can someone tell me what I am doing wrong?

Thanks