Solution
= Solution
The <chain rule> gives
$$
f'(x)=e^{x^3-x^5}(3x^2-5x^4)
=e^{x^3-x^5}x^2(3-5x^2).
$$
The <exponential function> never vanishes, so all <critical points> are
$$
\boxed{x=0,\qquad x=\sqrt{3/5},\qquad x=-\sqrt{3/5}.}
$$
In particular $f'(0)=0$.