I have to prove that
$$\vec{S}(\vec{r}) = -\frac{c^2}{\omega} \bigg( u(\vec{r})\vec{\nabla}\phi(\vec{r}) + \frac{i}{2}\vec{\nabla}u(\vec{r}) \bigg) ~~~~(1)$$
given that $$\vec{S}(\vec{r}) = \frac{-i}{\mu_0 \omega } \vec{E}(\vec{r}) \times \bigg(\vec{\nabla} \times \vec{E}(\vec{r})^* \bigg) ~~~ (2)$$ and $$\vec{E}(\vec{r}) = \hat \epsilon \sqrt{\frac{u(\vec{r})}{\epsilon_0}} e^{i \phi(\vec{r})} ~~~~ (3)$$ as well as $$\vec{\nabla} \times \vec{E}(\vec{r}) = 0 ~~~~ (4)$$ where $\hat \epsilon^2 = 1$ and $\epsilon_0 = \frac{1}{\mu_0 c^2}$ ($c =$ speed of light).
And I am not quite able to show that. What I've done so far is applying BAC-CAB to (3) ending up with
$$\vec{S}(\vec{r}) = - \frac{i}{\mu_0 \omega} \bigg[ \vec{\nabla} \bigg(\vec{E}(\vec{r}) \vec{E}(\vec{r})^*\bigg) - \vec{E}(\vec{r})^*\bigg(\vec{E}(\vec{r}) \vec{\nabla} \bigg) \bigg]$$
The left term simplifies to
$$ -\frac{i}{\mu_0 \omega} \bigg[ \vec{\nabla} \bigg(\vec{E}(\vec{r}) \vec{E}(\vec{r})^*\bigg)\bigg] = -\frac{i}{\mu_0 \omega} \bigg[ \vec{\nabla} \bigg(\Im (\vec{E}(\vec{r}))^2 + \Re(\vec{E}(\vec{r}))^2\bigg)\bigg] = -\frac{i}{\mu_0 \omega \epsilon_0} \bigg[ \vec{\nabla} \bigg( u(\vec{r}) cos^2(\phi(\vec{r})) + u(\vec{r}) sin^2(\phi(\vec{r})) \bigg) \bigg] = - \frac{i c^2}{\omega} \vec{\nabla} u(\vec{r}) $$
I've used that if $z = x + iy$ then $z\cdot z^* = x^2 + y^2$ where $\Im(z) = x$ and $\Re(z) = y$ as well as $e^{i\phi} = cos(\phi) + i sin(\phi)$.
For the second term however I am kind of clueless. I would write
$$\vec{E}(\vec{r})^*\bigg(\vec{E}(\vec{r}) \vec{\nabla}\bigg) = \bigg(\vec{E}(\vec{r}) \vec{\nabla}\bigg)\vec{E}(\vec{r})^* = \vec{E}(\vec{r}) \vec{\nabla}\vec{E}(\vec{r})^* $$
for being able to apply the nabla operator - but I don't see any sense in actually applying it...
What would be possible next steps to take?
Thank you very much for your help.
FunkyPeanut