How to compute the MLE of the zero truncated poisson? $$P(X = x ) =  
\frac{\theta ^ x e^{- \theta} }{x ! \left ( 1 - e^{- \theta} \right )},\ x=1,2,\cdots,\ 0<\theta<\infty$$
Then the likelihood of $(x_1,\cdots,x_n)$ is 
$$\mathcal L(\theta \mid \boldsymbol y) = \prod_{i=1}^n \frac{e^{-\theta}}{1-e^{-\theta}} \frac{\theta^{x_i}}{x_i!} \propto (e^{\theta} - 1)^{-n} \theta^{n \bar x}$$
so the log-likelihood is
$$\ell(\theta \mid \boldsymbol x) = -n \log(e^{\theta} - 1) + n\bar x \log \theta$$
and its derivative with respect to θ is
$$\frac{\partial \ell}{\partial \theta} = \frac{e^\theta}{1-e^{\theta}} + \frac{\bar x}{\theta}$$
but I don't know how to compute $\partial \ell/\partial\theta = 0$, since this does not have a closed form expression.
Is there anyone can give me an idea?

Your calculation is right so far. As you said it does not exist a closed form. We can use the Lambert W function to obtain a solution for $\theta$
$\frac{e^\theta}{1-e^{\theta}} + \frac{\bar x}{\theta}=0$
Multiplying both sides by $(1-e^{\theta})$ and $\theta$
$\theta\cdot e^\theta+\bar x\cdot (1-e^{\theta})=0$
$e^\theta \cdot (\theta-\bar x)+\bar x=0$
$e^\theta \cdot (\theta-\bar x)=-\bar x$
Multiplying both sides by $e^{-\bar x}$
$e^{\theta-\bar x} \cdot (\theta-\bar x)=-\bar x\cdot e^{-\bar x}$
${\theta-\bar x}=W(-\bar x\cdot e^{-\bar x})$
${\hat \theta}=W(-\bar x\cdot e^{-\bar x})+\bar x$
