 Solution

The second equilibrium point equation gives $y=-(x+1)/2$. Substituting into the first gives $x+e^{(x-1)/2}=0$, so $x<0$. Put $x=-a$, with $a>0$. ThenThe function $h(a)=a{e}^{(a+1)/2}$ is strictly increasing on $a\ge 0$, since $h^{\prime }(a)=e^{(a+1)/2}(1+a/2)>0$. It starts at zero and tends to infinity, so it takes the value one exactly once. Moreover $h(1)=e>1$, so $0<a<1$. This proves uniqueness of the stated equilibrium point.

At that point $e^{x+y}=a$, and the Jacobian matrix isIts trace is $3+a$ and determinant $2+a$. The characteristic discriminant is $(3+a{)}^2-4(2+a)=(a+1{)}^2$, so its two eigenvalues areThey are distinct and positive, with eigenvectors $(1,-1)$ and $(a,1)$ respectively. Thus the linearized critical point is an unstable node. It is a hyperbolic equilibrium, so the original nonlinear equilibrium point is also locally repelling; this last observation is stronger than the requested linearized classification.