I am trying to find the MLE for $\{X_i\}_{i=1}^{n}$ that have the following pdf
$$\frac{3x^2}{\theta^3}I(0\le x \le \theta)$$
I tried solving it directly by taking the log of the likelihood and then through the derivative solve for $\theta$, but the indicator function gives me issues. Intuitively, to maximize the mle I need a very small $\theta$ by the restriction $X_{(1)}$ is the MLE. Is this correct or is there a way to solve this directly?
You are correct in saying that you need the smallest $\theta$, but for the given problem, the smallest $\theta$ is going to be $X_{(n)}$ as opposed to $X_{(1)}$ since you want $\theta$ to be greater than all the $X_i$.
More concretely, the likelihood here is given by (assuming iid) \begin{equation*} \mathcal{L}(\theta) = \prod_{i=1}^n \frac{3x_i^2}{\theta^3}I(0\leq x_i \leq \theta) = \frac{3^n \prod_{i=1}^n x_i^2}{\theta^{3n}} I (0\leq x_{(n)} \leq \theta). \end{equation*} This is maximized when $\theta = x_{(n)}$.