Answer
\begin{align*}\frac{d^2y}{dt^2}&=
\left(4 t^{2}-4 t-1\right) e^{t-t^{2}}.
\end{align*}
Work Step by Step
Since we have
$$ y= e^{t-t^2} $$
then the first derivative $\frac{dy}{dt}$, using the chain rule, is given by
\begin{align*}
\frac{dy}{dt}&= e^{t-t^2} (t-t^2)'\\
&=(1-2t)e^{t-t^2}
\end{align*}
The second derivative $\frac{d^2y}{dt^2}$, using the product rule, is given by
\begin{align*}\frac{d^2y}{dt^2}&=(1-2t)'e^{t-t^2}+(1-2t)(e^{t-t^2})' \\
&=(-2)e^{t-t^2}+(1-2t)(1-2t)e^{t-t^2}\\
&=
\left(4 t^{2}-4 t-1\right) e^{t-t^{2}}.
\end{align*}