Maximal open set for implicit function theorem We consider the equation $\cos (xy) + \sin(xy) = y$ with $(x,y)\in\mathbb{R}^2$.
From the implicit function theorem, one has the existence of an open neighborhood of $0$ on which there exists a unique solution $y : x\mapsto y(x)$ to the equation.
I would like to know how to determine a maximal open set on which the map $y$ is defined. The hint proposes the use a drawing. But I do not know how to draw the implicit plot of the equation.
 A: We are told to study the equation
$$f(x,y):=\cos(xy)+\sin(xy)-y=0\tag{1}$$
in the neighborhood of an initial point $(0,y)$ satisfying the equation.
Since $f(0,y)=0$ implies $y=1$ there is exactly one such point, namely ${\bf z}_0:=(0,1)$. The point ${\bf z}_0$ is marked in the following contour plot of $f$.

One computes
$$f_x=y\bigl(-\sin(xy)+\cos(xy)\bigr),\quad f_y=x\bigl(-\sin(xy)+\cos(xy)\bigr)-1\ .$$
As $f_y({\bf z}_0)=-1\ne0$ the implicit function theorem allows to conclude the following: There are a window $W=[-a,a]\times[1-b,1+b]$ centered at ${\bf z}_0$ and a $C^1$-function
$$\phi:\quad [-a,a]\to[1-b,1+b],\qquad x\mapsto y=\phi(x)\ ,$$ 
such that within $W$ the equation $(1)$ is equivalent with $y=\phi(x)$, in particular $\phi(0)=1$. Furthermore one has
$$\phi'(0)=-{f_x({\bf z}_0)\over f_y({\bf z}_0)}=1\ .$$
The graph $\gamma$ of this $\phi$ is shown in red in the above figure. 
Now this implicit function statement does not tell us anything about the size of this window $W$, nor about what's happening with  $\phi$ outside of $W$. In order to gain global information about $\phi$ we have to invoke other tools. From the figure we conjecture that $\phi$ is actually defined for $-\infty<x<\infty$. In the following I shall give a parametric representation of the "full" $\gamma$.
Using the formula $\cos\alpha+\sin\alpha=\sqrt{2}\sin\bigl(\alpha+{\pi\over4}\bigr)$ we write the equation $(1)$ in the form
$$\sqrt{2}\sin\left(xy+{\pi\over4}\right)=y\ .\tag{2}$$
Looking at the above picture we get the impression that along $\gamma$ the variable $y$ grows from $0$ to $\sqrt{2}$ and then descends again to $0$. Looking at $(2)$ this suggests to introduce the auxiliary variable $$u:=xy+{\pi\over4}\qquad(0<u<\pi)\ ,$$
so that $y=\sqrt{2}\sin u$. 
This leads to the "trial"  parametric representation
$$\gamma_?:\quad u\mapsto\left\{\eqalign{x(u)&={u-\pi/4\over\sqrt{2}\sin u}\cr y(u)&=\sqrt{2}\sin u \cr}\right.\qquad(0<u<\pi)\ .\tag{3}$$
One easily checks that $$\lim_{u\to0+} x(u)=-\infty, \quad \lim_{u\to\pi-}x(u)=\infty,\qquad x'(u)>0\quad(0<u<\pi)\ ,\tag{4}$$
and above all, that $$f\bigl(x(u),y(u)\bigr)\equiv0\quad(0<u<\pi), \quad \gamma_?(\pi/4)={\bf z}_0\ .$$
The uniqueness part of the implicit function theorem then guarantees that within that window $W$ the curve $\gamma_?$ coincides with the graph of $\phi$; hence $\gamma_?$ is indeed the red curve $\gamma$ in the first figure. But $\gamma_? (=\gamma)$ can be considered as a graph of some function $\phi$ over all its length: From $(4)$ it follows that the function $u\mapsto x(u)$ $(0<u<\pi)$ has an inverse $\psi: \ x\mapsto u=\psi(x)$ $(-\infty<x<\infty)$. It follows that $\gamma_?$ can be viewed as graph of the function
$$\phi(x)=\sqrt{2}\sin\bigl(\psi(x)\bigr)\qquad(-\infty<x<\infty)\ ,$$
 I'm closing with a picture of the curve $\gamma_?$, as defined in $(3)$.

