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We consider the equation $cos (xy) + sin(xy) = y$ with $(x,y)inmathbb{R}^2$.

From the implicit function theorem, one has the existence of an open neighborhood of $0$ on which there exists a unique solution $y : xmapsto y(x)$ 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.

We are told to study the equation
$$f(x,y):=cos(xy)+sin(xy)-y=0tag{1}$$
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
$$f_x=ybigl(-sin(xy)+cos(xy)bigr),quad f_y=xbigl(-sin(xy)+cos(xy)bigr)-1 .$$
As $f_y({bf z}_0)=-1ne0$ 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
$$phi:quad [-a,a]to[1-b,1+b],qquad xmapsto y=phi(x) ,$$
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 $cosalpha+sinalpha=sqrt{2}sinbigl(alpha+{piover4}bigr)$ we write the equation $(1)$ in the form
$$sqrt{2}sinleft(xy+{piover4}right)=y .tag{2}$$
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 $$u:=xy+{piover4}qquad(0<u<pi) ,$$
so that $y=sqrt{2}sin u$.
This leads to the "trial" parametric representation
$$gamma_?:quad umapstoleft{eqalign{x(u)&={u-pi/4oversqrt{2}sin u}cr y(u)&=sqrt{2}sin u cr}right.qquad(0<u<pi) .tag{3}$$
One easily checks that $$lim_{uto0+} x(u)=-infty, quad lim_{utopi-}x(u)=infty,qquad x'(u)>0quad(0<u<pi) ,tag{4}$$
and above all, that $$fbigl(x(u),y(u)bigr)equiv0quad(0<u<pi), quad gamma_?(pi/4)={bf z}_0 .$$
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 $umapsto x(u)$ $(0<u<pi)$ has an inverse $psi: xmapsto u=psi(x)$ $(-infty<x<infty)$. It follows that $gamma_?$ can be viewed as graph of the function
$$phi(x)=sqrt{2}sinbigl(psi(x)bigr)qquad(-infty<x<infty) ,$$
I'm closing with a picture of the curve $gamma_?$, as defined in $(3)$.