Let $X \sim \text{Beta}_{(\theta, 1)} =: \mathbb{P}_\theta$ be a continuous random variable where $$\mathbb{f}_\theta(x) := \theta \cdot x^{\theta-1}\mathbb{1}_{[0,1]} = \cases{\theta \cdot x^{\theta-1} \hspace{10pt} \text{x}\in [0,1]\\0 \hspace{33pt} \text{x}\in [0,1]}$$
Calculate the distribution of the random variable $Y := -\log(X)$. My first attempt looks like this:
Let $h(x) := -\log(x)$. We get $h'(x) = -\frac{1}{x}$ and $h^{-1}(x) = e^{-x}$. Now I apply the density transformation:
$$\mathbb{f}^Y(y) = \frac{\mathbb{f}^X(h^{-1}(y))}{|\det(h'(h^{-1}(y)))|} = e^{-y} \cdot \theta \cdot {(e^{-y})}^{\theta-1}\cdot\mathbb{1}_{[0,1]}(y) = \theta e^{-\theta \cdot y}\cdot\mathbb{1}_{[0,1]}(y)$$
This result doesn't seem right to me, since it's integral is not equal to $1$, but I can't seem to find a mistake.
Your notation is somewhat confusing, let me denote by $f_X()$ and $f_Y()$ the densities of $X,Y$
Then , if $Y=h(X)$ and $h$ is monotonic and differentiable, we have
$$\begin{align} f_Y(y)&=\left.{\frac{f_X(x)}{|h'(x)|}}\right|_{x=h^{-1}(y)}\\ &= \left. \frac{\theta \cdot x^{\theta-1} }{1/x} [0\le x\le 1] \right|_{x=h^{-1}(y)}\\ &= \left. \theta x^{\theta} \, [0\le x\le 1] \right|_{x=\exp(-y)} \\ &= \theta e^{- \theta y } \, [0\le e^{-y}\le 1] \\ &= \theta e^{- \theta y } \, [y\ge 0] \\ \end{align} $$
which is an exponential.
Your error was in doing the change of variable within the indicator function. Here, I'm using the Iverson bracket notation instead.