I'm reading a textbook where it forms a Lagrangian function
$$ L = \int_0^1 f(x)^{1 - \frac{1}{\alpha}}dx - \lambda\int_0^1 g(x) f(x) dx$$ But how do you take the derivative of this thing? The variable that we want to optimize is $f(x)$, which is a function ($g(x)$ is a known function). So we need to take the derivative with respect to a function? How?
So in this case, what are the first-order conditions?
An important result of the calculus of variations is that if you have a functional $L[f(x)]$ such that $$ L[f] =\int_a^b J(x,f,f') dx, $$ $L$ is minimized if $$ \frac{\partial J}{\partial f} - \frac{d}{dx} \frac{\partial J}{\partial L'} = 0. $$ This is the Euler-Lagrange equation. In your case, $$ L[f] = \int_0^1 \left[f(x)^{1-1/\alpha}-\lambda g(x) f(x) \right] dx, $$ therefore the Euler-Lagrange equation leads to $$ \frac{\partial}{\partial f} \left[f(x)^{1-1/\alpha}-\lambda g(x) f(x) \right] - \frac{d}{dx} \frac{\partial}{\partial f'} \left[f(x)^{1-1/\alpha}-\lambda g(x) f(x) \right] =0. $$ The derivative in relation to $f'$ vanishes because $L[f]$ does not depend on $f'$. Evaluating the derivative in relation to $f$: $$ \frac{\alpha-1}{\alpha}f(x) ^{-1/\alpha} - \lambda g(x) = 0 $$ and now solving for $f$: $$ f(x) = \left( \frac{\lambda \alpha}{\alpha-1} g(x)\right)^{-\alpha}, $$ which is the $f(x)$ that minimize your $L$.