= Solution
Introduce one stage $k=f(t_n+h/2,Y)$ with $Y=y_n+hk/2$, and update $y_{n+1}=y_n+hk$. This is the <implicit midpoint rule> as a one-stage <Runge-Kutta method>, with <Butcher tableau>
$$
\begin{array}{c|c}1/2&1/2\\\hline&1\end{array}.
$$
For a smooth vector field, write $\Delta=y_{n+1}-y_n$. Expanding its stage equation gives
$$
\Delta=hf(t_n,y_n)+\frac{h^2}{2}\bigl(f_t+f_yf\bigr)(t_n,y_n)+O(h^3),
$$
which matches the exact <Taylor expansion> through degree two. The method therefore has order at least two. On the <Dahlquist test equation> $y'=\lambda y$, its <stability function> is
$$
R(z)=\frac{1+z/2}{1-z/2}=1+z+\frac{z^2}{2}+\frac{z^3}{4}+O(z^4),\qquad z=h\lambda.
$$
The third coefficient differs from the exponential's $1/6$, so the order is exactly two.
There is no pole in $\operatorname{Re}z\leq0$, and
$$
|1-z/2|^2-|1+z/2|^2=-2\operatorname{Re}z\geq0.
$$
Thus $|R(z)|\leq1$ throughout the closed left half-plane:
$$
\boxed{\text{one-stage implicit Runge-Kutta; order }2;\quad\text{A-stable}.}
$$
This conclusion is <A-stability>, not strong damping of arbitrarily stiff modes: $R(z)\to-1$ along the negative real axis.
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