(The likelihood function in the question is written mistakenly, the derivative of the log-likelihood is correct).
To address the specific issue raised by the OP, we need, as mentioned in the comments, to determine whether the second derivative is negative evaluated at the MLE. The tip here is that checking this does not necessarily require to obtain a closed-form expression for the MLE -in fact, sometimes, even if we have a closed-form expression, it is advisable not to use it, but rather, exploit the first-order conditions in a different way.
And this is exactly the case here. The first-order condition eventually leads to a second-degree polynomial in $\theta$ (including the sample mean of the data as a parameter), with a single admissible root, So we have a closed-form expression for the MLE. Denoting the sample mean by $\bar x$, this expression here is
$$\hat \theta = \frac {\bar x-1 + \sqrt{\bar x^2 +6\bar x +1}}{4} >0 \tag{1}$$
Let's turn now to the issue of the sign of the 2nd derivative. The second derivative is
$$\frac{d^2logL(\theta;x)}{d\theta^2}=-n\left(\frac{2\theta+2\theta^2-(2\theta+1)^2}{\theta^2(1+\theta)^2}\right) - 2\frac{\sum X_i}{\theta^3}$$
$$\Rightarrow \frac 1n \frac{d^2logL(\theta;x)}{d\theta^2}=-\frac 1{\theta^2}\left[\frac{-2\theta^2-2\theta-1}{(1+\theta)^2} + 2\frac{\bar x}{\theta}\right] \tag{2}$$
We want the term inside the brackets in $(2)$ to be positive. Now imagine inserting $(1)$, the closed-form expression for the MLE, into $(2)$, and try to determine the sign of the exrpession...
...Instead, it is more efficient to go back to the first-order condition and eliminate the sample mean, since the MLE must satisfy:
$$\bar x = \frac{(2\hat \theta+1)\hat \theta}{(1+\hat \theta)} \tag{3}$$
Inserting $(3)$ into the bracketed expression in $(2)$ we obtain
$$[\;] = \frac{-2\hat \theta^2-2\hat \theta-1}{(1+\hat \theta)^2} + 2\frac{1}{\hat \theta}\frac{(2\hat \theta+1)\hat \theta}{(1+\hat \theta)} $$
which for sign purposes is equivalent to
$$\frac{-2\hat \theta^2-2\hat \theta-1}{(1+\hat \theta)} + 2(2\hat \theta+1) >0$$
$$\Rightarrow (4\hat \theta+2)(1+\hat \theta) > 2\hat \theta^2+2\hat \theta+1$$
which holds, since $\hat \theta >0$.