We have three points labeled 1, 2 and 3 and an angle in the following image.
It is easy to show that the gradient of $\theta$ with respect to the position of the point 1 (i.e. when the other two points are fixed in space) is $$ \boldsymbol{\nabla}_1 \theta = \frac{1}{r_{12}} \hat{\boldsymbol{\mathrm{u}}}_1,$$ where $r_{12}$ is the magnitude of the vector $\boldsymbol{\mathrm{r}}_{12} \equiv \boldsymbol{\mathrm{r}}_1 - \boldsymbol{\mathrm{r}}_2$, and $\hat{\boldsymbol{\mathrm{u}}}_1$ is the unit vector perpendicular to $\boldsymbol{\mathrm{r}}_{12}$ and in the plane that includes the three points. Similarly, the gradient of $\theta$ with respect to the position of the point 3 can be written as $$ \boldsymbol{\nabla}_3 \theta = \frac{1}{r_{32}} \hat{\boldsymbol{\mathrm{u}}}_3.$$ Now, I'm looking for an expression for the gradient of $\theta$ with respect to the position of the point 2. Based on some insight from physics, I guess that the relation $\boldsymbol{\nabla}_1 \theta + \boldsymbol{\nabla}_2 \theta + \boldsymbol{\nabla}_3 \theta = 0$ holds here, though I haven't proved it mathematically.
Based on the relation
$$\cos(\theta)=\frac{\boldsymbol{\mathrm{r}}_{12}\cdot\boldsymbol{\mathrm{r}}_{32}}{r_{12}r_{32}},$$
I just derived the following relations:
$$\boldsymbol{\nabla}_1 \theta = \frac{1}{r_{12}} \frac{\cos(\theta)\hat{\boldsymbol{\mathrm{r}}}_{12}-\hat{\boldsymbol{\mathrm{r}}}_{32}}{\sin(\theta)},$$
$$\boldsymbol{\nabla}_3 \theta = \frac{1}{r_{32}} \frac{\cos(\theta)\hat{\boldsymbol{\mathrm{r}}}_{32}-\hat{\boldsymbol{\mathrm{r}}}_{12}}{\sin(\theta)},$$ and
$$\boldsymbol{\nabla}_1 \theta + \boldsymbol{\nabla}_2 \theta + \boldsymbol{\nabla}_3 \theta = 0.$$
Boldface letters with a hat ($\hat{}$) are unit vectors.