Asymptotic stability of fixed point $f'(t)=af(t)(K-f(t))-bf(t)g(t)$ for $a,b,c,d,t,K>0$
$$g'(t)=cf(t)g(t)-dg(t)$$
This system has 3 fixed points (You can evaluate them if you set the 2 equations = 0). One point is $(\frac{d}{c},\frac{a}{b}(K-\frac{d}{c}))$
I would like to know if this point is asymptotically stable for $K>\frac{d}{c}$, so if the solution converges to this point for $t\to\infty$, correct ?
I have no idea and would really appreciate if someone could show me how to do it so I can use the method for similar equations. 
 A: Consider a differential equation $X' = F(X)$, where $X(t) = (x_1(t),x_2(t))$.  This can also be written as
$$
\begin{align*}
x_1' &= F_1(x_1,x_2), \\
x_2' &= F_2(x_1,x_2),
\end{align*}
$$
where $F = (F_1,F_2)$.  Suppose the system has an equilibrium at $X = X_0$.  The first step is to compute the linearization about $X_0$, which is given by
$$
Y' = D_XF(X_0)\,Y.
$$
To unravel this notation a bit, suppose $X_0 = (\alpha,\beta)$.  Then $$D_XF(X_0) = \left(
\begin{array}{cc}
\frac{\partial F_1}{\partial x_1}(\alpha,\beta) & \frac{\partial F_1}{\partial x_2}(\alpha,\beta) \\
\frac{\partial F_2}{\partial x_1}(\alpha,\beta) & \frac{\partial F_2}{\partial x_2}(\alpha,\beta)
\end{array} \right).$$  This is just the matrix of the first partial derivatives of the components of $F$ evaluated at the equilibrium point.  In your problem, $X_0 = \left(\frac{d}{c},\frac{a}{b}\left(K-\frac{d}{c}\right)\right)$ and $$F(f,g) = \left(\begin{array}{c}
af(K-f)-bfg \\
cfg-dg
\end{array}\right)$$ (so that $F_1(f,g) = af(K-f)-bfg$ and $F_2(f,g) = cfg-dg$).
A linear system like this is asymptotically stable at the origin (and hence $X=X_0$ is an asymptotically stable equilibrium of $X' = F(X)$) if both eigenvalues of the matrix $D_XF(X_0)$ have negative real part.
For your problem, it is indeed the case that both eigenvalues have negative real part when $K > d/c$, so the equilibrium in question is indeed asymptotically stable.
