My functional is $J[f] = \int_{-\infty}^{\infty} f(x) \log f(x)\,dx$. I want to maximize it using the calculus of variations.
In order to use the Euler-Lagrange equation, I define $L(t, y, y')$ such that $J[f] = \int_{-\infty}^{\infty} L(t, f(x), f'(x))\,dt$:
$$L(t, y, y') = y \log y.$$
The EL equation says that
$$\frac{\partial}{\partial y} L(t, y, y') = \frac{d}{dt}\frac{\partial}{\partial y'} L(t, y, y').$$
(auxiliary question: why does the statement of the RHS of the EL equation use $\frac{d}{dt}$ and not $\frac{\partial}{\partial t}$?)
Substituting into the EL equation, the trivial solution is $\log y + 1 = 0$, or $f(x) = \exp(-1)$. I feel like I am missing something here. What's wrong?
I don't believe you are missing anything here. The expression above you have is for entropy and the distribution for maximal entropy on the reals is "uniform". In other words, the outcome of the experiment can be any value without any preference amongst them. You will however find that there is no such "probability density" since you can't integrate a positive value over the entire reals and obtain something finite.