Using Thevenin's Theorem?
It's entirely possible for both the short circuit current and open-circuit voltage to be zero. For example, if the network consists of a single resistor (or probably any linear network with no independent sources). 

In this case you just need to use a different pair of "test points" to find the equivalent resistance. For example, you can use V=0 (the short-circuit case) and V = 1 V. Then \$R_{th}=\frac{1}{I(V=1) - I(V=0)}\$, or more generally \$R_{th}=\frac{V_1-V_2}{I(V_1) - I(V_2)}\$.

Basically one way to state Thevenin's theorem is that the I-V characteristic of any one-port network of independent sources and linear elements will be a straight line. And \$R_{th}\$ is the slope of that line, \$\dfrac{\mathrm{d}v}{\mathrm{d}i}\$. Just as with any other linear function, you can can use any two points to find the slope of the line as rise/run.

Edit


  You would find \$V_{OC}\$ (voltage across the short circuit) and \$I_{SC}\$, the source current, and the Thevenin resistance would be \$R_{Th} = \frac{V_{OC}}{I_{SC}}\$.


You've mis-stated this a bit (where I added emphasis).

The usual way to find the Thevenin or Norton equivalent circuit is to find \$V_{OC}\$, the voltage across an open-circuit on the output, and \$I_{SC}\$, the current through a short-circuit on the output.

It happens that \$V_{OC}\$ is also the Thevenin-equivalent voltage source value and \$I_{SC}\$ is the Norton-equivalent current source value. But when you're measuring an arbitrary network to find the equivalents, you should think of these as the open-circuit output voltage and short-circuit output current.


  But is it possibly for both the voltage across the [open] circuit and the  [short-circuit] current to be zero, leaving the Thevenin resistance indeterminate?


I believe \$V_{OC}\$ and \$I_{SC}\$  will both be zero whenever there are no non-zero independent sources in the network being modeled. However this does not mean that the Thevenin resistance is indeterminate, as I explained above.

You could also contrive a case where there's a non-zero independent source and a dependent source that directly opposes it, like for example,







simulate this circuit &ndash; Schematic created using CircuitLab


  Can circuits with dependent and independent have indeterminate Thevenin resistances?


I think this is a different question. The Thevenin equivalent resistance will only be indeterminate if it goes to infinity, for example when the network being modeled is an ideal current source.