(∂_t) N1 = r N1(1-(N1/K))-a N1 N2 (1-Exp[-b N1])
(∂_t) N2 = -d N2+N2 e (1-Exp[-b N1])

(r N1(1-(N1/k))-a N1 N2 (1-Exp[-b N1]))/(k r)/. N1-> u  k/. N2 ->  v r/a/. t ->  τ/r /. e -> α r /. d -> δ r /. b -> β/k//FullSimplify
(-d N2+N2 e (1-Exp[-b N1]))/(r^2/a)/. N1-> u  k/. N2 ->  v r/a/. t ->  τ/r /. e -> α r /. d -> δ r /. b -> β/k//FullSimplify

f1[u_,v_]:=u-u^2-u v(1-Exp[-u β])
f2[U_,v_]:= v α(1-Exp[-u β])-v δ

Zeroes[f1[u,v]]

Solve[f1[u,v]==0,v]
Solve[f1[u,v]==0,u]

So, we want either of the following two solutions:

u=0

v=(((E^(u β)) (1-u))/(-1+(E^(u β))))

Solve[f2[u,v]==0,v]
Solve[f2[u,v]==0,u]//Simplify

{{u->-((Log[((α-δ)/( α))])/β)}}

And, we want either

v=0

u->-((Log[1-(δ/α)])/β)

Here, we evaluate the v term for the two different possible values of u

(((E^(u β)) (1-u))/(-1+(E^(u β))))/. u-> -((Log[1-(δ/α)])/β)//FullSimplify
(((E^(u β)) (1-u))/(-1+(E^(u β))))/. u->0 //FullSimplify

So, we have the following steady state solutions:

(0,0)
(0,∞)
(-((Log[1-(δ/α)])/β),0)
(-((Log[1-(δ/α)])/β),((α (β+Log[1-(δ/α)]))/(β δ)))

Now, calculate A

A = {{(∂_u) f1[u,v],(∂_v) f1[u,v]},{(∂_u) f2[u,v],(∂_v) f2[u,v]}}//FullSimplify

A//MatrixForm

Eigensystem[A/.u-> 0/. v-> 0]
Eigensystem[A/.u-> -((Log[1-(δ/α)])/β)/. v-> 0]//FullSimplify
Eigenvalues[A/.u-> -((Log[1-(δ/α)])/β)/. v->((α (β+Log[1-(δ/α)]))/(β δ))]//FullSimplify

{-(1/(2 β δ))(δ Log[1-(δ/α)] (-1+β+Log[1-(δ/α)])-α Log[1-(δ/α)] (β+Log[1-(δ/α)])+(Sqrt[Log[1-(δ/α)] (4 β (α-δ) (δ^2) (β+Log[1-(δ/α)])+Log[1-(δ/α)] (((α β+δ-β δ+(α-δ) Log[1-(δ/α)]))^2))])),(1/(2 β δ))(-δ Log[1-(δ/α)] (-1+β+Log[1-(δ/α)])+α Log[1-(δ/α)] (β+Log[1-(δ/α)])+(Sqrt[Log[1-(δ/α)] (4 β (α-δ) (δ^2) (β+Log[1-(δ/α)])+Log[1-(δ/α)] (((α β+δ-β δ+(α-δ) Log[1-(δ/α)]))^2))]))}/. β-> -Log[1-δ/α]//FullSimplify

FullSimplify[{(1/2) (-1+((δ Log[1-(δ/α)])/(Sqrt[(δ^2) ((Log[1-(δ/α)])^2)]))),-(1/2)-((δ Log[1-(δ/α)])/(2 (Sqrt[(δ^2) ((Log[1-(δ/α)])^2)])))},Assumptions ->( δ/α)<1&&δ>0&&α>0&&δ∈Reals&&α∈Reals]