Problem. Find the vertical asymptotes of the graph $(\!{\rm C}\!)$ of the function $y= \dfrac{1}{x\,\sin\frac{1}{x}}$ .
Solution. All the solutions of the equation $x\,\sin\dfrac{1}{x}= 0$ are $x= 0$ or $x= \dfrac{1}{w\pi}\,(\!w\in \mathbb{Z}\!)$ .
Therefore, we can see that the possible vertical asymptotes are $x= 0$ or $x= \dfrac{1}{w\pi}\,(\!w\in \mathbb{Z}\!)$ .
The line $x= 0$ cannot be a vertical asymptote of $(\!{\rm C}\!)$. Consider the following sequence $x_{n}= \dfrac{1}{\frac{\pi}{2}+ 2\,n\pi}$. This sequence tends to $0^{\!+}$, and the corresponding sequence $$y_{n}= \frac{\frac{\pi}{2}+ 2\,n\pi}{\sin\left ( \dfrac{\pi}{2}+ 2\,n\pi \right )}\rightarrow +\!\infty$$ On the other hand, if we choose $x_{n}= \dfrac{1}{\frac{-\!\pi}{2}\!+\!2\,n\pi}$, which also tends to $0^{\!+}$, the corresponding sequence $$y_{n}= \frac{- \frac{\pi}{2}+ 2\,n\pi}{\sin\left ( - \dfrac{\pi}{2}+ 2\,n\pi \right )}\rightarrow -\!\infty$$ Hence the limit does not exist when $x$ tends to $0$ on the right. Similarly for the left. Now we prove that all the remain lines are actually the vertical asymptotes of the graph $(\!{\rm C}\!)$ .
Let consider the case $x= \dfrac{1}{\pi}$. We have $$\lim_{x\rightarrow \left ( \dfrac{1}{\pi} \right )^{\!+}} y= \lim_{x\rightarrow \left ( \dfrac{1}{\pi} \right )^{\!+}} \frac{\pi}{\sin \left ( \dfrac{1}{x} \right )}\rightarrow +\!\infty$$ Notice that the denominator tends to $0$ but is always positive and therefore the limit is positive infinity. Similarly for the others vertical lines. In conclusion, all the vertical asymptotes of $(\!{\rm C}\!)$ are $$x= \dfrac{1}{w\pi}\,(\!w\in \mathbb{Z}\!)$$ I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !