Answer
$$y'=2\theta e^{-\theta^2}\sin(e^{-\theta^2})$$
Work Step by Step
$$y=\cos(e^{-\theta^2})$$
According to the Chain Rule, we have $$y=-\sin(e^{-\theta^2})(e^{-\theta^2})'=-\sin(e^{-\theta^2})(e^{-\theta^2})(-\theta^2)'$$
$$y'=-e^{-\theta^2}\sin(e^{-\theta^2})(-2\theta)$$
$$y'=2\theta e^{-\theta^2}\sin(e^{-\theta^2})$$