= Solution
Let $p_1,p_2$ denote the incoming $\phi,\Phi$ <four-momenta> and $p_3,p_4$ the outgoing $\phi,\Phi$ <four-momenta>. Thus $p_1+p_2=p_3+p_4$, $p_1^2=p_3^2=m^2$ and $p_2^2=p_4^2=M^2$. Define the <Mandelstam variables> and exchanged <four-momenta> by
$$
s=(p_1+p_2)^2=q_s^2,\qquad u=(p_1-p_4)^2=q_u^2,
\qquad q_s=p_1+p_2,\quad q_u=p_1-p_4=p_3-p_2.
$$
The leading connected four-point <scattering amplitude> uses two cubic <interaction vertices>. There are four <tree-level Feynman diagrams>: $s$ and $u$ exchange of $\Phi$ with two $\lambda$ vertices, and $s$ and $u$ exchange of $\phi$ with two $g$ vertices.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2019-mixed-scalar-tree-channels.png]
{title=The four tree channels for mixed-scalar scattering. Blue solid lines represent $\phi$ and orange dashed lines $\Phi$. External momenta are incoming on the left and outgoing on the right; crossings have no vertex.}
There is no term proportional to $\lambda g$: one vertex of each kind has three field slots of each species altogether, whereas the external state uses two of each. This leaves one $\phi$ and one $\Phi$ to contract internally, and their mixed <scalar propagator> is zero. There is also no $t$ channel: grouping both external $\phi$ legs at one vertex would require a $g$ vertex with an internal $\Phi$, while grouping both external $\Phi$ legs at the other would require a $\lambda$ vertex with an internal $\phi$.
Define $i\mathcal M$ after removing the overall <momentum conservation> delta function. The nonderivative exchange gives
$$
i\mathcal M_{\lambda^2}=(i\lambda)^2
\left[\frac{i}{s-M^2+i0}+\frac{i}{u-M^2+i0}\right].
$$
For the derivative $s$ channel, the incoming $\phi$ <four-momenta> at the first vertex are $p_1,-q_s$, and at the second they are $q_s,-p_3$. The two <interaction vertices> are therefore $ig\,p_1\cdot q_s$ and $ig\,p_3\cdot q_s$. The same reasoning with $q_u$ applies to the $u$ channel, giving
$$
i\mathcal M_{g^2}=-ig^2\left[
\frac{(p_1\cdot q_s)(p_3\cdot q_s)}{s-m^2+i0}
+\frac{(p_1\cdot q_u)(p_3\cdot q_u)}{u-m^2+i0}\right].
$$
The <mass shell> conditions and <momentum conservation> imply
$$
p_1\cdot q_s=p_3\cdot q_s=\frac{s+m^2-M^2}{2},\qquad
p_1\cdot q_u=p_3\cdot q_u=\frac{u+m^2-M^2}{2}.
$$
Combining the four <Feynman diagrams> therefore yields
$$
\boxed{\begin{aligned}
\mathcal M={}&-\lambda^2\left[\frac1{s-M^2+i0}+\frac1{u-M^2+i0}\right]\\
&-\frac{g^2}{4}\left[
\frac{(s+m^2-M^2)^2}{s-m^2+i0}
+\frac{(u+m^2-M^2)^2}{u-m^2+i0}\right].
\end{aligned}}
$$
Each displayed <tree-level Feynman diagram> has <Feynman-diagram symmetry factor> 1 for fixed external species and labels. The result is of order $\lambda^2,g^2$; if both couplings are scaled by a common parameter $\varepsilon$, the next connected corrections begin at order $\varepsilon^4$. A four-point cubic <Feynman diagram> has $3V=2I+4$, so its vertex count $V$ is even. The internal lines remain off shell throughout: imposing an on-shell relation on an exchanged <four-momentum> would incorrectly change the derivative vertices.
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