Background
The finite difference Euler method is
\begin{eqnarray} y_{n+1} &=& T(y_n,t_n)\\ &=& y_{n} + hf(y_n,t_n). \end{eqnarray}
Then for some perturbation $\delta$
\begin{eqnarray} y_{n+1} + \delta_{n+1} &=& T(y_{n} + \delta_{n},t_n)\\ &=& y_{n} + 2hf(y_n,t_n) + \delta_{n}\frac{\partial T(y_n,t_n)}{\partial y_n} \end{eqnarray}
giving a growth factor of
\begin{eqnarray} g_{n} &=& \frac{\partial T(y_n,t_n)}{\partial y_n} \end{eqnarray}
where stability requires $|g_n| < 1$.
For second order Runge-Kutta (RK2) methods we have
\begin{eqnarray} y_{n+1/2} &=& \frac{h}{2}f(y_n,t_n),\\ y_{n+1} &=& hf(y_{n+1/2},t_{n+1/2})\\[10pt] g_n &=& \frac{\partial y_{n+1/2}}{\partial y_n}\\ &=& 1 + h\frac{\partial f(y_{n+1/2},t_{n+1/2})}{\partial y_{n+1/2}}\bigg[1+\frac{h}{2}\frac{\partial f(y_n,t_n)}{\partial y_n}\bigg] \end{eqnarray}
For a small $h$ the below assumption gives growth in terms of the complex plane $\Delta$
\begin{eqnarray} \Delta &=& h\frac{\partial f(y_n,t_n)}{\partial y_n}\\ &\approx& h\frac{\partial f(y_{n+1/2},t_{n+1/2})}{\partial y_{n+1/2}}\\[10pt] g_n &=& 1 + \Delta + \frac{\Delta^2}{2}. \end{eqnarray}
Question
For the leap frog method
\begin{eqnarray} y_{n+1} &=& y_{n-1} + 2hf(y_n,t_n) \end{eqnarray}
w.r.t. the complex plane $\Delta$, why is the stability
\begin{eqnarray} g_n^2 &=& 1 + 2\Delta g_n? \end{eqnarray}
I got the following
\begin{eqnarray} g_n &=& \frac{\partial y_{n+1}}{\partial y_n}\\ &=& \frac{\partial y_{n-1}}{\partial y_n} + 2h\frac{\partial f(y_{n},t_n)}{\partial y_n}\\ &=& 1 + 2h\frac{\partial f(y_{n},t_n)}{\partial y_n}\\[10pt] \Delta &=& h\frac{\partial f(y_{n},t_n)}{\partial y_n}\\[10pt] \Rightarrow g_n &=& 1 + 2\Delta. \end{eqnarray}
I am unsure where the extra $g_n$ comes from on either side, as multiplying both sides by $g_n$ would give $g_n^2 = g_n + 2\Delta g_n$.