Differentiating between a square wave or sawtooth wave with a circuit...?
If the frequency for both waves is going to be 100 kHz with the same amplitude, you could construct a narrow bandpass filter at 200 kHz to put the signal through.  In theory a pure squarewave should only have odd harmonics, so there should not be much output at the second harmonic frequency.  On the other hand, a sawtooth wave has booth even and odd harmonics, so you will get a greater output.  The peak amplitude for the second harmonic of a sawtooth wave will simply be \$ \frac{2A}{\pi} \$, where A is the peak amplitude of the input sawtooth.  If wish you can then follow up the output of the bandpass filter with a peak detector and some kind of comparator.