Let's say you are given $\omega_f \in \mathcal{S}'(\mathbb R)$ with \begin{align*} f \colon \mathbb{R} &\to \mathbb{R}\\ x &\mapsto x, \end{align*} and the definition of Fourier transform
\begin{align*} \hat{\phi}(\omega)=\int_{-\infty}^{+\infty}e^{-i\omega x}f(x)dx, \, \omega \in \mathbb{R}. \end{align*} Obviously, that function has no Fourier transform, but its corresponding tempered distribution $\omega_f$ does. Using \begin{align*} \widehat{ (\partial_j \phi)} (\omega) =& \,\, i\omega_j\hat{\phi}(\omega)\,\,\,\,\,\,\,\,\,\, (\star) \\ \widehat{1} =& \,\, 2\pi \delta(\omega) \end{align*} we find \begin{equation} \widehat{ (\partial_jf)}(\omega)= \widehat{1}\Rightarrow\hat{f}(\omega)=-2 \pi i \frac{\delta(\omega)}{\omega}. \end{equation} On the other hand, I know that \begin{align*} 2\pi\delta(\omega)=\int_{-\infty}^{\infty}e^{-i\omega x}dx \Rightarrow 2\pi \delta'(\omega)=-i\underbrace{\int_{-\infty}^{\infty}xe^{-i\omega x}dx}_{=:\,\hat{f}(\omega)} \end{align*} which gives \begin{equation} \hat{f}(\omega) = 2\pi i\delta'(\omega). \end{equation}
Why do this two results contradict eachother?
Is it because ($\star$) is not true for $\phi \in \mathcal{S}'(\mathbb R^n)$?
Writing $\frac{\delta(\omega)}{\omega}$ is not a good idea, because strictly speaking, it's not well defined in many cases. But we do have the relation $\omega\delta'(\omega)=-\delta(\omega)$, so in this sense, your two answers are the same, they are both correct.
P.S. As for your second solution, it is helpful to think that way, and it can give you the correct answer quickly, but this is not how maths works.