I often see precision measurements of the $t\bar{t}H$ process at the (HL-)LHC justified by the statement that "it's the only cross-section that's directly proportional to the Higgs-top Yukawa coupling".
What's the cross-section $\sigma_{t\bar{t}H}$? I'm looking for the full expression, rather than just "$\sigma\sim y_t^2$".
(optional) Is that statement true?
Note: I mean production of a top quark pair in association with a Higgs boson, as in the diagram below.
If all you want to understand is why $\sigma$ is proportional to the top Yukawa then the answer is pretty easy. Cross-sections are proportional the square of the amplitude for the process. Higgs production with associated tops has two amplitudes, the one in your question and one that looks exactly like it but with the gluons swapped.
To calculate an amplitude from a diagram you use Feynman rules. Each external line comes with some kinematic function of an outgoing particle's momentum, each internal line comes with a propagator, and each vertex in the diagram carries a coupling constant. The coupling constant for the vertex between two top quarks and the Higgs boson is -- by definition -- $y_t/\sqrt{2}$.
Ultimately the amplitude is $$\mathcal{M} = \frac{g_s^2y_t}{\sqrt{2}}(T^aT^b)_{xy}\epsilon_\mu(k_1)\epsilon_\nu(k_2)\,\,\overline{u}(p_{t1})\gamma^\mu \frac{1}{\underline{p}_{t1}-\underline{k}_1-m_t}\frac{1}{-\underline{p}_{t2}+\underline{k}_2-m_t}\gamma^\nu v(p_{t2})+(a\leftrightarrow b,k_1\leftrightarrow k_2)$$ where the only dependence on $y_t$ is the explicit factor on the left (and the fact that $m_t=y_tv/\sqrt{2}$, but that's a small effect).
Since $\mathcal{M}\propto y_t$, $\sigma_{pp\rightarrow t\overline{t}H}\propto y_t^2$. This is not the only process that is proportional to $y_t$, as gluon-gluon fusion is also dominated by a diagram that is proportional to $y_t$.
