= Solution
The <derivative> is $f'(x)=-4x^3e^{-x^4}$. Its absolute value is <continuous> and tends to zero as $x\to\pm\infty$, so it is bounded. More explicitly,
$$
\sup_{x\in\mathbb R}|f'(x)|=4\left(\frac34\right)^{3/4}e^{-3/4}=:M,
$$
obtained by differentiating $4t^3e^{-t^4}$ for $t\geq0$. The <mean value theorem> gives $|f(x)-f(y)|\leq M|x-y|$ for every $x,y$. Thus $f$ is <Lipschitz continuous>, and choosing $\delta=\varepsilon/M$ proves \b[$e^{-x^4}$ is uniformly continuous on $\mathbb R$].
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