Here is the problem statement:
Let $(X_i,Y_i)$ be a random sample from a distribution with pdf $$f(x,y;\theta)= > \frac{1}{\theta^3}\exp\left(\frac{-x}{\theta}-\frac{y}{\theta^2}\right)\qquad > 0<x, 0<y.$$ (a) Find the MLE for $\theta$.
(b) Give the asymptotic distribution of $\sqrt{n}(\hat{\theta}-\theta).$
Here are my thoughts: We basically have two random samples, $X\sim \exp(\theta)$ and $Y\sim \exp(\theta^2)$. This pdf is the joint pdf for $X$ and $Y$. The parameter has no impact on the support of the variables $X$ and $Y$, so calculus should suffice for this problem. $$L(x,y;\theta) = \prod_{i=1}^n f(x,y;\theta) = \frac{1}{\theta^{3n}}\exp\left(\frac{-1}{\theta^2}\sum_{i=1}^n \theta x_i+y_i\right),$$ so $$l(x,y;\theta) = -3n\ln(\theta)-\frac{1}{\theta^2}\sum_{i=1}^n \theta x_i + y_i,$$ which has partial $\theta$ derivative $$ \frac{-3n\theta^2+\sum_{i=1}^n \theta x_i +2y_i}{\theta^3}.$$ Equating to zero and solving for $\theta$ requires solving a quadratic in $\theta$.
The second $\theta$ derivative is $$\frac{3n\theta^2 - \left(\sum_{i=1}^n 2\theta x_i + 6y_i\right)}{\theta^4}.$$ The solutions to the quadratic are $$\theta = \frac{\sum_{i=1}^n x_i\pm \sqrt{\left(\sum_{i=1}^nx_i\right)^2+24n\sum_{i=1}^n y_i}}{6n}.$$ Is there no easier way to do the problem? Am I forced to make this substitution to find the maximum?