Except that the integral is probably a limit, the idea is ok.
We need to show that the preimage of an open set is an open set.
Set ##f(x,y)=y\sin\left(\dfrac{x}{y}\right)## and ##U(p,r)=\{q\in \mathbb{R}^n\,|\,\|q-p\|<r\}.## An open set in the domain around ##1## contains an open ball ##U(1,\varepsilon)=\{p\in \mathbb{R}\,|\,\|p-1\|<\varepsilon\},## and we want to show that ##f^{-1}\left(U(1,\varepsilon)\right)=\{(x,y)\in \mathbb{R}^2\,|\,f(x,y)\in U(1,\varepsilon)\}## is open. We therefore show that it contains an open ball ##U((\pi/3,2),\delta).## This means that we have to show that ##f\left(U((\pi/3,2),\delta)\right)\subseteq U(1,\varepsilon).##
Now we choose a point ##(x,y)\in U((\pi/3,2),\delta)## which means that $$\|(x,y)-(\pi/3,2)\|=\sqrt{(x-\pi/3)^2+(y-2)^2} <\delta\quad(1)$$
is given, and we have to show that
$$
\|f(x,y)-1\|=\left|y\sin\left(\dfrac{x}{y}\right)-1\right|<\varepsilon\quad (2).
$$
So we have to show the last inequality by using the first. However, I do not follow your inequalities. Yes, we may assume that ##|y|>1,## but this has to be noted for our choice of ##\delta## or must be proven. And then, the first inequality is ##\le ## and not strict, since the sine term could be zero. Regardless of those tiny mistakes, I do not see where the ##-1## and the sine in the next steps go.
The problem is to show that ##x/y## is near ##\pi/6## only using the Euclidean distance. I would show that ##f## is continuous, partially differentiable, and continuity follows from that. I have tried to use the series expansion, but with no luck. One has to get rid of ##x/y## first.