How to prove that $F(x,y)=x+y^2+\sin(xy)$ defines implicitly a continuously differentiable function?

Question

I need help with this problem:

Let $F(x,y)=x+y^2+\sin(xy)$. Proce that in a sufficiently small neighbourhood of $(0,0)$ the equation $F(x,y)=0$ defines implicitly a continuously differentiable function $g$ such that $g(0)=0$ and $F(x,y)=0$ if and only if $x=g(y)$. Show also that $g'(0)=0$. Apply theorem 4.7.14 with the roles of x and y reversed.

I don't know how to begin, I first tried to prove it like this: $$F(0,0)=0$$ $$\frac{\partial F}{\partial x}(0,0)=1\neq0$$ Thus, the exists a neighbourhood $N$ of $0$, a neighbourhood $M$ for $0$ and a continuously differentiable function $g:N\subset\mathbb{R}\rightarrow\mathbb{R}$, such that $g(0)=0$, and for each $x\in N$, $F(x,y)=0$ is uniquely solved by $x=g(y)$. Am I right? What should I do next?

Maybe you can give me some hints or guidance.

Answer 1

That's great so far, the next step is to show $g'(0) = 0$.

Our theorem tells us that $g'(0) = -\frac{\partial F}{\partial y}|_{0,0}/\frac{\partial F}{\partial x}|_{0,0} = -\frac{0}{1} = 0$.

We have swapped the partial fractions around, in the theorem you have linked it is $f(x) = y$, but we have $g(y) = x$. So our $g'$ is given by $\partial_y F / \partial_x F$.