Well, the title says it all. I need to prove that if $(X,Y)$ are jointly normal and $\mathrm{e}^X$ and $\mathrm{e}^Y$ are uncorrelated, then so are $X$ and $Y$.
Although the book (A Probability Path, by Resnick) does not make it explicit, I'm assuming that "jointly normal" means a bivariate normal distribution (as follows from here and here, subsection "Bivariate case").
I really don't know how to start solving this. Any help or hint will be appreciated. Thank you very much in advance.
Thanks to the hint given by @JGWang I could solve this. I provide my answer here:
Let $(X,Y) \sim \mathcal{N}((\mu_X, \mu_Y), \Sigma)$, with $$\Sigma = \begin{pmatrix} \sigma_X^2 & \mathrm{Cov}(X,Y) \\\mathrm{Cov}(Y,X) & \sigma_Y^2 \end{pmatrix}.$$
By completing the square, we can prove that $$\mathbf{E}(\mathrm{e}^X) = \exp\left(\frac{(\mu_X + \sigma_X^2)^2 - \mu_X^2}{2\sigma_X^2}\right),$$ idem with $Y$. Now, if we denote $\rho = \frac{\mathrm{Cov}(X,Y)}{\sigma_Y\sigma_Y}$, then it is known that the density function of $(X,Y)$ is \begin{align} f_{X,Y}(x,y) =& \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}} \times \dotsb\\ & \dotsb \times\exp \left(-\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_X)^2}{\sigma_X^2} + \frac{(y-\mu_Y)^2}{\sigma_Y^2} - \frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y}\right]\right). \end{align} With this, we can replicate the computations made before (i.e. by completing the square) and prove that $$\mathbf{E}(\mathrm{e}^{X+Y}) = \exp\left(\mu_X + \mu_Y + \frac{1}{2}\left( \sigma_X^2 + \sigma_Y^2 + 2\rho \sigma_X\sigma_Y \right)\right).$$ Given that $\mathrm{e}^X$ and $\mathrm{e}^Y$ are uncorrelated, then we have that $\mathbf{E}(\mathrm{e}^{X+Y})=\mathbf{E}(\mathrm{e}^{X})\mathbf{E}(\mathrm{e}^{Y})$. Matching terms, we get $$\rho \sigma_X \sigma_Y = 0 \implies \rho = 0 \implies \mathrm{Cov}(X,Y) = 0,$$ where both implications are true because $\sigma_X,\sigma_Y > 0$.