Calculus, water poured into a cone: Why is the derivative non-linear?

I understand that $\frac {dh}{dt}$ is decreasing. However, I don't understand on an intuitive level why $\frac {dh}{dt}$ is non-linear.

Because any strictly decreasing function that is linear eventually becomes negative, but you already know that $\frac {dh}{dt}$ is always positive.

The volume of water is changing linearly, but the height and volume are related nonlinearly. That is why $h(t)$ is non-linear.

\begin{eqnarray*} V &=& \frac{1}{3} \pi r^2 h\\ r &=& h \tan \theta\\ V &=& \frac{1}{3} \pi \tan^2 \theta h^3\\ \frac{dV}{dt} &=& \frac{1}{3} \pi \tan^2 \theta 3 h^2 \frac{dh}{dt}\\ \end{eqnarray*}

$\frac{dV}{dt}$ is a constant because the constant filling rate. So you can see from the above equation that $\frac{dh}{dt}$ is not constant.

The top radius $r$ of the cone is proportional to $h$: We have $r(t)=c\>h(t)$ for some constant $c$. Therefore $V(t)=c\> h^3(t)$ with some other $c$, or $h(t)=c\> V^{1/3}(t)$. This implies $${dh\over dt}=c\> V^{-2/3}(t) V'(t)=c\> V^{-2/3}(t)\ ,$$ since $V'(t)$ is constant. Now $t\mapsto V(t)$ is linear; hence $t\mapsto V^{-2/3}(t)$ is a "root function", and not linear.