Electronic – How to determine if an edge-triggered flip-flop is positive going or negative going

Yes, such a thing is possible, but it isn't more useful than more conventional flip-flops. Here's one way to make one:

^{simulate this circuit – Schematic created using CircuitLab}

The output is the XOR of two internal flip-flops. If they're different, the output is high; otherwise, the output is low.

REG1 will toggle on the rising edge of the S input only if the output is low; if the output is high, it won't change state.

Similarly, REG2 will toggle on the rising edge of the R input only if the output is high; if the output is low, it won't change state.

Note that there's a requirement that near-simultaneous rising edges on both S and R have a minimum spacing, basically determined by the delay time of the feedback gates.

The only change that is required to convert your falling-edge triggered flip-flop to a rising-edge triggered flip-flop is to swap the true (non-inverted) clock and the inverted clock at the pins of your tri-state buffers and the transmission gate. For example, the center transmission gate in a rising-edge triggered flip-flop would have the true clock connected to the NMOS transistor and the inverted clock connected to the PMOS transistor.

By the way, near the middle left of your diagram you have a circled inverter that shows Q at its output and !Q at its input. That's correct...the output of the inverter just to the right of the circled inverter is !Q (Q-bar) rather than Q. If you follow the path from the D input through an inverting tristate buffer and the NOR gate you will see that the signal at the right end of the (non-inverting) transmission gate must have the same polarity as D, so that point in your circuit represents the true (non-inverted) Q value.