If vector field $F$ is $\langle P, Q, R \rangle$, where $P$, $Q$ and $R$ are each functions of $x$, $y$ and $z$, $curl F$ is:
$\langle (\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}), (\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}), (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}) \rangle$
Now $curl (curl F)$ would be:
$\langle (\frac{\partial (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})}{\partial y}-\frac{\partial (\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})}{\partial z}), (\frac{\partial (\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})}{\partial z}-\frac{\partial (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})}{\partial x}), (\frac{\partial (\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})}{\partial x}-\frac{\partial (\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})}{\partial y}) \rangle$
I think I can use the chain rule to simplify that, but I'm not sure how.
You can't really simplify it much, but you can use the fact that the derivative of a difference is the difference of derivatives. For example: $$\dfrac{\partial\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)}{\partial x}=\dfrac{\partial}{\partial x}\dfrac{\partial Q}{\partial x}-\dfrac{\partial}{\partial x}\dfrac{\partial P}{\partial y}=\dfrac{\partial^2 Q}{\partial x^2}-\dfrac{\partial^2 P}{\partial x\partial y}$$