Let $(X_i, Y_i), i = 1,...,n$, be independent and identically distributed random variables. The pdf for $x > 0$ and $y > 0$ is given by $f_{\theta} (x,y) = e^{-\theta x -y/\theta}$, where $\theta > 0$ is an unknown parameter.
How do I find the MLE of $\theta$? The part that makes it confusing for me, is that this is a bivariate pdf.
It doesn't change nothing substantial, because when you derive the MLE you are considering the likelihood function over the parametric space $\Theta$ where the sample points $\{x_i, y_i\}$ are already known. As such, it doesn't matter how many variables in your multivariate original distribution, as it only matters how many parameters you have to estimate, i.e., the dimension of your parametric space rather than the number of variables. \begin{align} L(\theta;x_1,...,x_n,y_1,...,y_n)=\prod e^{-\theta x_i}e^{-y_i/\theta}=e^{\theta \sum x_i}e^{-\theta^{-1} \sum y_i}, \end{align} so $$ l(\theta)=-\theta\sum x_i - \sum y_i/\theta, $$ $$ l'(\theta)=-\sum x_i+\sum y_i /\theta^2 = 0, $$ $$ \hat{\theta}_n = \left( \frac{\sum y_i}{\sum x_i} \right)^{1/2}. $$ Reassuring that the point is indeed minimum, $$ l''(\hat{\theta}) = -\sum y_i/\hat{\theta}^3 < 0. $$