I am trying to differentiate this equation.
$$y = \frac{At}{Bt^2 + Ct^3}$$
Is this correct?
$$y' = \frac{(Bt^2 + Ct^3) \cdot A - At(2Bt + 3Ct^2)}{(Bt^2 + Ct^3)^2}$$
$$= \frac{ABt^2 + ACt^3 - 2ABt^2 - 3ACt^3}{(Bt^2 + Ct^3)^2}$$
$$= \frac{-ABt^2 - 2ACt^3}{(Bt^2 + Ct^3)^2}$$
Is this as simplified as this equation can get?
So far your derivative is all correct, though you have some more algebraic simplification that you can do:
$$\frac{-ABt^2 - 2ACt^3}{(Bt^2 + Ct^3)^2}$$
Let's first factor the numerator.
$$-ABt^2 - 2ACt^3 = -At^2(B+2Ct)$$
And now the denominator.
$$(Bt^2+Ct^3)^2 = (t^2(B+Ct))^2 = t^4(B+Ct)^2 $$
We've got the $t^2$ like term that we can reduce:
$$\frac{-At^2(B+2Ct)}{t^4(B+Ct)^2} = \frac{-A(B+2Ct)}{t^2(B+Ct)^2}$$
We have no more ways to reduce like terms, so the final reduced answer is:
$$ -\frac{A(B+2Ct)}{t^2(B+Ct)^2} $$