How can I prove that $\int _{-1}^{1} \frac{1}{x} dx =0 $

This cannot be proven rigorously because it is not technically true.

The Cauchy principal value of your integral is $0$, but if you go back to the definition of the improper Riemann integral you will find that this quantity is undefined.

The integral is zero when we consider the principal value of the integral. I assume that this is what Wolfram Alpha confirms.

To prove this, split the integral up over its positive and negative parts and replace the zero with a variable and have it tend towards zero.$$\begin{align*}\int\limits_{-1}^1\frac {\mathrm dx}x & =\lim\limits_{\varepsilon\to0}\left[\int\limits_{-1}^{-\varepsilon}\frac {\mathrm dx}x+\int\limits_{\varepsilon}^1\frac {\mathrm dx}x\right]\\ & =\lim\limits_{\varepsilon\to0}\Bigr[\log|-\varepsilon|-\log|-1|+\log 1-\log(\varepsilon)\Bigr]\\ & =\lim\limits_{\varepsilon\to0}\log\left(\frac {\varepsilon}{1}\frac 1{\varepsilon}\right)\\ & =\log 1\\ & =0\end{align*}$$