Question about Implicit function theorem I was asked a simple question, show that $y+\sin y=x$ sets in the neighborhood of $(0,0)$ $y$ as a function of $x$, and find $\dfrac{dy}{dx}(0,0)$
Firstly, my naive solution would be:
Since $lim_{y \to 0} \frac{\sin y}{y} = 1$ I want to say that in the vicinity of $(0,0)$ $y=\sin y$ and then we get the equation $2y=x$, $y=\dfrac{1}{2}x$, and so $\dfrac{dy}{dx}(0,0)=\dfrac{1}{2}$
But if I was to do this question using the implicit fucntion theorem I have a problem.
Let's define $f(x,y)=y+\sin y-x$
Let's look at the matrix of the partial derivatives: $\begin{pmatrix} \dfrac{df}{dx}(0,0) & \dfrac{df}{dy}(0,0)\end{pmatrix} = \begin{pmatrix} -1 & 2 \end{pmatrix}$ and of course this matrix is not invertible, it is not even a square matrix.
So my questions are:
1) Why is my naive solution wrong?
2) Why isn't the jacobi matrix a square matrix?
3) Even if it was an invertible matrix, how would I find $y$ as a function of $x$ in order to find $\dfrac{dy}{dx}$?
 A: I'll start by answering 2.
In the notation used in the wikipedia link you have $m=n=1$. So the jacobian matrix is a $1\times 2$ matrix. It isn't square simply because $n\neq 0$.
In 3, it's not the jacobian matrix that needs to be invertible, but rather the matrix $\begin{pmatrix}\dfrac{\mathrm df}{\mathrm dy}(0,0)\end{pmatrix}$ and it is. 
Therefore there exist intervals $V,W$ around $0$ such that "$y\colon V\to W$ is a smooth function of $x$", that is there exists a smooth function $g\colon V\to W$ such that $\forall x\in V\exists !y\in W(g(x)=y\land f(x,y)=0)$.
Define $h\colon V\to W, x\mapsto f(x,g(x))$. By differentiating, the chain rule yields $$\forall x\in U\left(0=h'(x)=\dfrac{\partial f}{\partial x}(x,g(x))+\dfrac{\partial f}{\partial y}(x,g(x))g'(x)\right).$$
In particular for $x=0$ one gets $g'(0)=-\dfrac{\dfrac{\partial f}{\partial x}(0,g(0))}{\dfrac{\partial f}{\partial y}(0,g(0))}=-\dfrac{\dfrac{\partial f}{\partial x}(0,0)}{\dfrac{\partial f}{\partial y}(0,0)}=\dfrac 1 2$.
A: Let $f(x,y)=y+\sin y-x$. So, $\frac{\partial f}{\partial y}(x,y)= 1+\cos y$ and $\frac{\partial f}{\partial x}(x,y)= -1$ are continuos functions. Then, $f(x,y)$ is $C^1$. 
Therefore, by Implicit function theorem, exists $I$x $J \in \mathbb{R}^2$ st 
$f^{-1}(0) \cap$ $(I$x $J)$ is a graphic of a function in $x$. By chain rule, we got:
$$y'(x)=-\frac{\partial f/ \ \partial x}{\partial f / \partial y } \Rightarrow y'(0)=1/2$$
