Example of invariance of Wald Test In order to make us understand that the Wald doesn't work with a transformation of your parameter at H0, we have to test the following $H_0$'s:
A) $H_0: exp( \mu)=0.62$
B) $H_0:  \mu=ln(0.62)$
I can write the Wald test statistic for case B, which is 
$W_B= (\mu - ln(0.62))^2 / Var( \mu)$ which matches the result we get out of Stata,
But what is the formula for $W_A$? 
I think you have to apply the delta method, but I can't figure it out unfortunately. Thank you in advance!
 A: The problem you face is that $\sigma^{2}_{g(\mu)} \ne g(\sigma^{2}_{\mu})$, that is, the variance of a function of a variable, is not the same thing as the same function of the variance of the variable.
The Delta function (Oehlert, 1992) gives us a way to approximate the variance of the function of a variable. When applying the delta function there is a trade-off between complexity (corresponding to the order of a Taylor series expansion entailed in the application) and accuracy of the approximation. However, the first order application of the delta method is often considered acceptable.
In general, the first order approximation looks like $g\left(X\right) \approx g\left(\mu\right) + g'\left(\mu\right)\left(X-\mu\right)$.
For the variance of a function, $g(\cdot)$, of the data, $X$ with $E(X)=\mu$, this translates into $\sigma^{2}_{g\left(X\right)} = \left[g'\left(\mu\right)\right]^{2}\sigma^{2}_{X}$.
If $g(X) = e^{X}$, then $\sigma^{2}_{e^{X}} = \sigma^{2}_{X}e^{2X}$.
So I think the Wald $\chi^{2} = \frac{e^{2X}}{\sigma^{2}_{X}e^{2X}}$, and the Wald $t = \frac{e^{X}}{\sqrt{\sigma^{2}_{X}e^{2X}}}$.

References
Oehlert, G. W. (1992). A note on the delta method. The American Statistician, 46(1):27–29.
A: This is how I understood it: 
We test the hypothesis that $exp(\theta_0) = exp(\hat{\theta})$. As we are applying a function to our variable, we have to use the delta method to calculate the variance: $\sigma_{g(\hat{\theta})}^2={[g'(\hat{\theta})]}^2\sigma_{\hat{\theta}}$, which is in our case equal to $\sigma_{\hat{\theta}}^{2} exp(2\hat{\theta})$ and hence our test statistic is equal to $\frac{{[g(\hat{\theta})-g(\theta_0)]}^2}{\sigma_{g(\hat{\theta})}^2}=\frac{{[exp(\hat{\theta})-exp(\theta_0)]}^2}{\sigma_{\hat{\theta}}^{2}exp(2\hat{\theta})}$ which is chi-squared distributed with one degree of freedom.
Unfortunately the calculation of this test statistic is not yet identical as the one of the stata output, so maybe I am still doing something wrong...
