Use Lyapunov-Schmidt reduction to find an expression, or approximation, of the set of equilibria, as a function of the parameter $\lambda$, of the planar vector field $$f(x,y,\lambda)=(\lambda + 2x + y - x^2, 2x + (1+\lambda)y - xy)$$ near the equilibrium $(x,y) = (0,0)$ at $\lambda = 0$.
So, usually, we tackle this problem by using the Implicit Function Theorem twice. Firstly, we define $f_1(x,y,\lambda) = \lambda + 2x + y - x^2, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $f_2(x,y,\lambda) =2x + (1+\lambda)y - xy$.
Since $f_1(0,0,0) = 0$ and $D_x f_1(0,0,0)=2$, so it's invertible, the implicit function theorem states that there exist open neighborhoods $U \subseteq \mathbb{R}$ and $V \subseteq \mathbb{R}^2$ of $0$ and a function $x^*: V \to U$ such that $f_1(x,y,\lambda) = 0 \iff x^*(y,\lambda) = x$, with $x^*(0,0) = 0$. Furthermore, $D_y x^*(0,0) = -(D_x f_1 (0,0,0))^{-1} D_y f_1 (0,0,0) = -(2)^{-1}1 = -\tfrac{1}{2}$.
Now, we substitute $x^*(y,\lambda)$ for $x$ in $f_2$, and again we want to apply the IFT to express $y$ in terms of $\lambda$. So we define $g(y,\lambda) = f_2(x^*(y,\lambda), y, \lambda) = 2x^*(y,\lambda) + (1+\lambda)y - x^*(y,\lambda)y$. However, $D_y g(0,0) = 2D_y x^*(0,0) + 1 + 0 - D_y x^*(0,0)\cdot 0 -x^*(0,0) =0$, so we can't apply the IFT.
Is this just an ill-chosen example, or is there a different approach possible?