Let's assign variables to pertinent values:
- The height of the cone-shaped coffee filter: $h$ (in cm)
- The radius of the cone-shaped coffee filter at a certain height $h$: $r$
- The rate at which the cone-shaped filter is being filled with coffee: $\frac{{dv}}{{dt}} = 10$ cm$^3/$s

We want to find $\frac{{dh}}{{dt}}$ when $h = 6$ cm.

From the given information, we can relate $h$ and $r$ using similar triangles. By the Pythagorean theorem, the radius of the cone-shaped filter at a height $h$ is given by:
\begin{align*}
r &= \frac{hr_{\text{{top}}}}{h_{\text{{top}}}}
\end{align*}where the radius of the top of the cone-shaped filter is $r_{\text{{top}}} = 4$ cm and the total height of the cone-shaped filter is $h_{\text{{top}}} = 12$ cm.

Differentiating implicitly with respect to $t$ (the constant being differentiated) and using the chain rule, we have:
\begin{align*}
\frac{{dr}}{{dt}} &= \frac{{d}}{{dt}}\left(\frac{hr_{\text{{top}}}}{h_{\text{{top}}}}\right) \\
\frac{{dr}}{{dt}} &= \frac{{h_{\text{{top}}}r_{\text{{top}}}\frac{{dh}}{{dt}} - hr_{\text{{top}}}\frac{{dh_{\text{{top}}}}}{{dt}}}}{{\left(h_{\text{{top}}}\right)^2}}
\end{align*}Since the rate at which the cone is being filled with coffee $\frac{{dv}}{{dt}} = 10$ cm$^3/$s, we can also relate the height and radius via the volume formula for a cone:
\begin{align*}
v &= \frac{1}{3}\pi r^2 h
\end{align*}Differentiating implicitly with respect to $t$ and using the chain rule, we have:
\begin{align*}
\frac{{dv}}{{dt}} &= \frac{{d}}{{dt}}\left(\frac{1}{3}\pi r^2 h\right) \\
10 &= \frac{1}{3}\pi\left(r^2\frac{{dh}}{{dt}} + 2rh\frac{{dr}}{{dt}}\right) \\
10 &= \frac{1}{3}\pi\left(6^2\frac{{dh}}{{dt}} + 2(6)(h)\frac{{dr}}{{dt}}\right) \\
10 &= 12\pi\left(\frac{{dh}}{{dt}} + h\frac{{dr}}{{dt}}\right)
\end{align*}At $h = 6$ cm, we want to find $\frac{{dh}}{{dt}}$. Given that $r_{\text{{top}}} = 4$ cm and $h_{\text{{top}}} = 12$ cm, we can determine $r$ at $h = 6$ cm:
\begin{align*}
r &= \frac{hr_{\text{{top}}}}{h_{\text{{top}}}} \\
r &= \frac{(6)(4)}{12} \\
r &= 2
\end{align*}Substituting the values we know into the equation we derived earlier, we have:
\begin{align*}
10 &= 12\pi\left(\frac{{dh}}{{dt}} + (6)\left[\frac{{hr_{\text{{top}}}}}{h_{\text{{top}}}}\right]\right) \\
10 &= 12\pi\left(\frac{{dh}}{{dt}} + (6)\left(\frac{(6)(4)}{12}\right)\right) \\
10 &= 12\pi\left(\frac{{dh}}{{dt}} + (6)(6)\right) \\
10 &= 12\pi\left(\frac{{dh}}{{dt}} + 36\right)
\end{align*}Dividing both sides by $12\pi$ and isolating $\frac{{dh}}{{dt}}$:
\begin{align*}
\frac{10}{12\pi} &= \frac{{dh}}{{dt}} + 36 \\
\frac{10}{12\pi} - 36 &= \frac{{dh}}{{dt}} \\
\frac{10}{12\pi} &= \frac{{dh}}{{dt}} - \frac{36}{1} \\
\frac{10}{12\pi} &= \frac{{dh}}{{dt}} - \frac{{36}{1}{12\pi}}{1}
\end{align*}Simplifying the right side:
\begin{align*}
\frac{10}{12\pi} &= \frac{{dh}}{{dt}} - \frac{{36}{1}{12\pi}}{1} \\
\frac{10}{12\pi} &= \frac{{dh}}{{dt}} - 36\pi
\end{align*}Then, substituting the value of $h = 6$ cm into the equation:
\begin{align*}
\frac{10}{12\pi} &= \frac{{dh}}{{dt}} - 36\pi \\
\frac{10}{12\pi} &= \frac{{dh}}{{dt}} - 36\pi \\
\end{align*}Solving for $\frac{{dh}}{{dt}}$:
\begin{align*}
\frac{{dh}}{{dt}} &= \frac{10}{12\pi} + 36\pi \\
\frac{{dh}}{{dt}} &= \frac{10}{12\pi} + \frac{{36}{12}{\pi}}{1} \\
\frac{{dh}}{{dt}} &= \frac{10 + 36\pi^2}{12\pi}
\end{align*}Thus, when the height of the coffee in the filter is 6 centimeters deep, the height of the coffee is increasing at a rate of $\boxed{\frac{10 + 36\pi^2}{12\pi}}$ cm/s.

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