Let $x=\tan\theta$ where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, then $$ \begin{array}{lrcl} &\frac{x}{\sqrt{x^2 +1}}+\frac{1-x^2}{1+x^2} &>& 0\\ \Leftrightarrow&\frac{\tan\theta}{\sqrt{\tan^2\theta+1}}+\frac{1-\tan^2\theta}{1+\tan^2}&>&0\\ \Leftrightarrow&\sin\theta+(\cos^2\theta-\sin^2\theta)&>& 0\\ \Leftrightarrow&\sin\theta+(1-2\sin^2\theta) &>&0\\ \Leftrightarrow&(2\sin\theta + 1)(\sin\theta -1)&<&0 \end{array} $$ Because $(\sin\theta- 1)\le 0$ always hold unless $\sin\theta = 1$, the above relation can hold if and only if $$2\sin\theta + 1 > 0\ \text{and}\ \sin\theta <1 \implies 1 > \sin\theta > - \frac{1}{2}\implies \frac{\pi}{2} > \theta > -\frac{\pi}{6}$$ Accordingly, we find $$\boxed{x > -\frac{\sqrt{3}}{3}}$$