12.24 The function is defined for all $x$. We find

$$
y^{\prime}=3 x^{2} e^{-x}-x^{3} e^{-x}=x^{2} e^{-x}(3-x)
$$

The equation $y^{\prime}=0$ has only one root $x=3$. Since $y^{\prime}>0$ for $x<3$ and $y^{\prime}<0$ for $x>3$, the function reaches a maximum at the point $x=3$, which is equal to $3^{3} e^{-3}=\frac{27}{e^{3}}$.

Answer: $\left(3 ; \frac{27}{e^{3}}\right)$ - the point of maximum.
$12.24 y=x^{3} e^{-x}$.