Consider the explicit midpoint method, i.e $$y_{n+1}-y_{n-1} = 2hf(y_n).$$
I'm asked to apply this method to the linear test equation, $f(y_n) = \lambda y_n,$ in order to find the method's stability region.
So I have $$y_{n+1}-y_{n-1} = 2h\lambda y_n.$$
In the problems I've done before I wouldn't have the $y_{n-1}$ term so I could just easily write $y_{n+1} = R(z)y_n.$ How would I solve this?
You have written down the midpoint rule as a two-step method, a member of the family of multi-step methods. For these methods, one can show that a multi-step method $$\alpha_{k} y_{n+k} + \alpha_{k-1} y_{n+k-1} + \ldots + \alpha_{0} y_{n} = h\left( \beta_{k} f_{n+k} + \ldots + \beta_{0} f_{n}\right)$$ is called stable if the polynomial $$\rho(z) = \alpha_{k} z^{k} + \alpha_{k-1} z^{k-1} + \ldots + \alpha_{0}$$ has only roots lying on or within the unit circle and that roots lying on the unit circle are simple roots.
So it is up to you to demonstrate for which stepsize $h$ the roots of the polynomial $\rho(z) = z^{2} - 2h z - z$ satisfy the above criterion.
The other approach would be as you hint through the test equation. And then, you should indeed re-write the method as a single-step method:$$y_{n+1} = y_{n} + h\lambda y_{n+\frac{1}{2}} = y_{n} + h\lambda \left(y_{n} + \frac{1}{2}h \lambda y_{n}\right) = y_n \left(1+h\lambda + \frac{1}{2}\left(h\lambda\right)^{2}\right)$$ and this gives you your $R(z)$ function as $$R(z) = 1+h\lambda + \frac{1}{2}\left(h\lambda\right)^{2}$$ so your stability criterion in $(x,y)$ form says: $$|1+x+\frac{x^2-y^2}{2}| \leq 1 \ \text{and}\ |y+xy| \leq 1$$
Solving this gives you a graph like below.
