Now I'm trying to prove the Sawtooth Ramp wave output of ADF4159 with tektronix oscilloscope as the below. But as you can see the below image, it's FREQUENCY-TIME graph. So I want to know how to prove the Sawtooth Ramp wave output of ADF4159 with tektronix oscilloscope. Would you please help me?
Note that your chip isn't going to make a 12 GHz signal on its own. You need to use it with an appropriate VCO, loop filter, and reference clock generator to make a complete synthesizer.
Having done that, you can pre-scale down your 12 GHz synth's output and use it as the REF input to a second PLL (probably optimized for fast tracking rather than low phase noise). Then use your 'scope to monitor the VCO control signal of the 2nd PLL to get the frequency-time graph you're looking for.
You possibly can't without some extra hardware or capabilities. Oscilloscopes aren't typically designed for frequency vs time. If your scope actually has a high enough sample frequency and bandwidth, it probably also has an fft mode in which you can view the signal in the frequency domain. But even then it will show you magnitude vs. freq and not freq vs time (and you would see something like a single rectangular pulse centered at the middle frequency).
Some fancy scopes have Matlab built in. In this case you could demodulate the incoming samples in real time, to extract the frequency component, and then output the freq signal to the oscilloscope viewport. That's how I would do it if I were required to.
What you're looking to do is plot a measurement trend of a frequency measurement.
This is built into some oscilloscopes (at Keysight, it's the 3000T X-Series oscilloscopes and up), but not others. Often, it's a math channel.
If your scope can't plot that, you're out of luck.
If you need to see a 12 GHz signal, you'll need a very high bandwidth oscilloscope (pronounced "expensive") and it will likely be able to do it.
I'd also point out that if you just need to check the frequency of a sine wave, you could probably get away with a lower bw scope, but the signal will look very attenuated.