Harmonic function and surface integrals. I am trying to give a solution to this question but I'm getting stuck.

Let $f$ be an harmonic function $V \to \mathbb R$ where $V$ is a solid in $\mathbb R^3$ bounded by a solid surface $S$ with normal $\vec n$. Show that
$$ \iint_S \frac{\partial f}{\partial n} \ dS = 0$$

For the first part I said that a harmonic function is a function that satisfies $$\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=0$$ I think this is right.
Then for the next part I tried to use the divergence theorem to say
$$\iint_S \frac{\partial f}{\partial n}dS=\iiint_R \text {div}(\frac{\partial f}{\partial n})dV$$ then I get stuck working out $$\text {div}(\frac{\partial f}{\partial n})$$
I know it is the dot product between $\nabla$ and $f$ but I don't know how to split $f$ into a vector to take the dot product considering I don't even know what $f$ is.
If anyone here could help guide me that would be fantastic.
 A: Hints:
This question is mostly about getting the definitions straight. Once that's done you should find the one 'computational' step relatively straight forward.
The integral $$ \iint_S \frac{\partial f}{\partial\vec n} \ dS$$ is, by definition of the directional derivative $\partial/\partial \vec n$, equal to
$$\iint_S \nabla f \cdot \vec n \ dS$$
Now apply the divergence theorem to turn this into an integral over the volume $V$. 
Note also that as $V$ is a region in $\mathbb R^3$, the Laplacian of $f$ is $$\nabla^2 f = f_{xx} + f_{yy} + f_{zz},$$ not just the first two terms as in your question.
A: Divergence theorem tells us that $$ \iiint_V div(F) \, dV = \iint_S F \cdot \vec{n} \, dS $$
Note that $\frac{\partial f}{\partial n} = \nabla{f} \cdot \vec{n}$ so 
$$ \iint_S \frac{\partial f}{\partial n} dS = \iint_S \nabla{f} \cdot \vec{n} \, dS = \iiint_V div(\nabla{f}) \, dV = \iiint_V \nabla^2{f} \, dV$$
The condition you've written for a harmonic function is a statement for a function of 2 variables of the general condition that $\nabla^2{f} = 0$ so the integrand is $0$ and hence the whole integral is $0$.
