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Let's use the substitution $u=e^x$. There are a few reasons to think this is a good choice. It's an “inside function," in that if we let $f(x)=e^x$, then $f(e^x)=e^{e^x}$, which is a piece of our integrand. Also its derivative, $e^x$, is multiplied by the rest of the integrand, since $e^{2x}=e^x\cdot e^x$ .

Let $u=e^x$, $\mathrm{d}u = e^x\mathrm{d}x$. When $x=0$, $u=1$, and when $x=1$, $u=e$.

$$\begin{aligned} \displaystyle\int_0^1e^{2x}e^{e^x}\mathrm{d}x &= \int_0^1e^{x}e^{e^x}e^x\mathrm{d}x =\int_1^e ue^u~\mathrm{d}u \\ \text{This is more familiar. We use integration by parts with $\mathrm{d}v = e^u~\mathrm{d}u$, $v=e^u$. Conveniently, the ``$u$" we brought in with the substitution is what we want to use for the ``$u$" in integration by parts, so we don't have to change the names of our variables.} \\ &=\big[ue^u\big]_1^e - \int_1^e e^u~\mathrm{d}u\\ &=e\cdot e^e-e-e^e+e=e^e(e-1) \end{aligned}$$