Show that the second derivative of $y=(x^2−3x)^3$ is $6(x^2-3x)(5x^2-15x+9)$ How do you find the second derivative of $$y=(^2−3)^3$$  and show that it is $6(x^2-3x)(5x^2-15x+9)$
Every time I worked it out, I used the chain rule first to get the first derivative which is
$ {3(x^2-3x)^3}\cdot{2x-3}$. Then I used the product rule + chain rule but I only managed to get the expanded form of what I was supposed to "show" which is $30x^4-180x^3+324x^2-162x$
Any help would be appreciated.

You did not get the first derivative correctly.
The first derivative is $3(x^{2}-3x)^{2}(2x-3)$ and the second derivaive is $6(x^{2}-3x) (2x-3)^{2}+ 6(x^{2}-3x)^{2}$. Factor out $(x^{2}-3x)$ to get $6(x^{2}-3x) (4x^{2}-12x+9+x^{2}-3x)=6(x^{2}-3x)  (5x^{2}-15x+9)$
