Finding a diode's operation point (Q-point)
I agree with your value for \$V_{th}\$, but not with your \$R_{th}\$. You have 50 &Omega; between the two diode nodes, and then a parallel network, so the equivalent resistance can never be greater than 50 &Omega;. I found 31.66 &Omega;.  

But that still means the point (3.36 V, 0 mA) is off the graph. Now you can calculate the line's equation, but that's not even necessary. If for 0 V you have 3.37 V/ 31.66 &Omega; = 10.6 mA, then moving halfway to the (3.37 V, 0 mA) point the line will drop to half its height, so (3.37/2 V, 10.6/2 mA) is also a point on the line, and that is on the graph, so you can draw the line between (0 V, 10.6 mA) and (1.68 V, 5.3 mA):



So the diode's operation point is at 625 mV, for a current of 8.8 mA.