We know the volume of the cone is $\displaystyle \frac{1}{3}\pi r^{2}h$. Now, we know that there must be a constant relationship between the sides of the right triangle formed by any height $\displaystyle h$ and any radius $\displaystyle r$, since the angle between the vertical (the height) of the cone and the side of the cone is always $\displaystyle \frac{\pi}{6}=30^{\circ}$. It's a 30-60-90 triangle! Here is a diagram of the triangle created:

$\displaystyle h=r\sqrt{3}$ or $\displaystyle r=\frac{h}{\sqrt{3}}$

Let's plug that into the cone volume equation:

$\displaystyle V=\frac{1}{9}\pi h^{3}$. Now differentiate with respect to $\displaystyle t$:

$\displaystyle \frac{dV}{dt}=\frac{1}{3}\pi h^{2}\frac{dh}{dt}$

$\displaystyle \frac{dV}{dt}=0.1$, so plug that in and just solve for $\displaystyle \frac{dh}{dt}$, which will be in terms of $\displaystyle h$.