Let $h\colon\mathbb{R}^2\to\mathbb{R}$ defined by $h(x,y)=x^2+y^3+xy+x^3+ay$ (where $a\in\mathbb{R}$).
a) Find the values of $a$ for which $h(x,y)=0$ defines $y$ as a $\mathscr{C}^1$ implicit function of $x$ in some open neighborhood $U\times V$ of $(0,0).$
b) Find the values of $a$ for which $h(x,y)=0$ defines $x$ as a $\mathscr{C}^1$ implicit function of $y$ in some open neighborhood of $(0,0)$.
c) Let $f$ be the implicit function $U\to\mathbb{R}$ found in question a), and let $F\colon U\times\mathbb{R}\to\mathbb{R}$ defined as $$F(x,y)=\big(e^{x+y}+x^2-1\mathbin,f(x)+y\cos(x)\big).$$ Show that $F$ has an inverse $F^{-1}$, of class $\mathscr{C}^1$, defined in some neighborhood of $(0,0)$. Show that $G:=F\circ F+F^{-1}$ is differentiable at $(0,0)$, and calculate ${\rm D}G(0,0)$.
Hi everyone, I'm having problems with this exercise, in the first part, what I did was using the implicit function theorem to show that if $a \neq 0$, then y=f(x) and bla, bla. BUT, if $a \neq 0$, I don't know how to procede, 'cause the differential isn't invertible, and that isn't enough to prove that x doesn't define y as an implicit function.
Same goes for the second one, cause the differential is equal to zero, and I can't use the implicit function theorem. So, any ideas of how I should proceed?
In question a), for $a=0,$ the equation becomes $$(y+x)(y^2-xy+x^2+x)=0$$ and has only one $C^1$ solution on a neighborhood of $0:$ $$x\mapsto y=-x$$ because the two other solutions $y$ are real only if $x(4+3x)<0,$ i.e. (for $x$ close to $0$) only if $x<0.$
In question b), for $a=0,$ besides the solution $$y\mapsto x=-y,$$ there is another one near $(0,0),$ which is also $C^1:$ $$y\mapsto x=\frac{y-1+\sqrt{1-2y-3y^2}}2.$$ But for $a\ne0,$ there is no solution $y\mapsto x$ mapping $0$ to $0$ and differentiable at $0,$ because it would require $x=O(y)$ hence (plugging this into the equation) $ay+O(y^2)=0,$ which is absurd.