Why is dp/dv zero for a real gas in critical conditions?
This is merely a shard of a fact which does not make much sense in and by itself. After all, in systems with gas/liquid equilibrium there is nothing really special about $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. On the contrary, this is pretty typical. See all those points where the blue lines (isotherms) are horizontal? They make up a whole huge area on the diagram.


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Indeed, if you have a liquid in equilibrium with its saturated vapor, then the pressure does not depend on volume (in an isothermal processes, that is). As you compress it a little more, some gas turns to liquid; as you release it, some liquid evaporates. The pressure still remains the same - namely, the saturated vapor pressure at given temperature. Things start changing only when you pull the piston all the way out or push it all the way in.

Now, the two-phase area, huge as it is, does not comprise all diagram. If you heat the gas past the critical temperature, you'll get an isotherm of a different sort (it is not shown on the picture, so just imagine one, positioned above the rest of them). It will be all smooth and monotonic, without those horizontal parts.

Well, then there must be a temperature where this behavior changes, and that's what we call the critical temperature. Sure enough, it produces a very special isotherm: there is only one point on it where $\left(\dfrac{\partial\mathfrak p}{\partial V}\right)_T=0$. (That's what we call the critical point, BTW.) Moreover, since it is an inflection point, it also has $\left(\dfrac{\partial^2\mathfrak p}{\partial V^2}\right)_T=0$. Now that's the fact one should remember about the critical point.

The peculiar behavior of pretty much all physical properties around the critical point gives rise to various weird phenomena like critical opalescence, but that's another story.

So it goes.