What do u get when u differentiate y= 1/[3-2(t^2)] and then differentiate again

What do you get when u do dy/dx and then d^2y/dx^2?
I need to check if my answer is correct. Thanks.

I suppose that you really mean dy/dt and d2y/dt2

y = 1/(3 - 2t^2) = (3 - 2t^2)^-1 ----> dy/dt = -1*(3 - 2t^2)^-2 * -4t

dy/dt = 4t / (3 - 2t^2)^2

d2y/dt2 = 4*(3 - 2t^2)^-2 + 4t* -2(3 - 2t^2)^-3 * -4t

= (12 + 24t^2) / (3 - 2t^2)^3